4^x + 3 = 7 using the change of base formula log y =
log y/log b

The statement that is true is: D. Randomly selected students were used.

Survey can be defined as the data or information that a researcher or research analyst collected from a population so as to reach or draw a conclusion.

Based on the scenario the student were randomly selected which means that the randomly selected students were used to conduct the survey.

Therefore the correct option is D.

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A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. An isosceles triangle is one with two equal-length sides.

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angle of a triangle is always equal to 180°.

In ΔBAD and ΔBDC,

m∠ADB = m∠CDB {Given}

BD = BD {Common side in two triangles}

AD ≅ DC {Given}

Using the ASA postulate the two triangles are congruent. Therefore,

m∠BAD= m∠BCD.........equation 1

m∠ABD= m∠CBD

In ΔADC, AD ≅ DC {Given}, therefore, the triangle is an isosceles triangle. Thus,

m∠DAE = m∠DCE.......equation 2

Adding two of the equation 1 and 2,

m∠BAD + m∠DAE = m∠DCE + m∠BCD

∠BAC = ∠BCA

Since in ΔABC, ∠BAC = ∠BCA therefore, the triangle is an isosceles triangle, thus, AB = BC,

BE is the common side between ΔAEB and ΔCEB

Also, m∠ABD= m∠CBD

Therefore, ΔAEB ≅ ΔCEB

Now as ΔAEB ≅ ΔCEB, therefore, AE=EC,

Thus, BE is the median of the isosceles triangle ABC, and for an isosceles triangle, the median opposite to the non-common sides is the perpendicular bisector to the opposite side.

Hence, BD⊥AC.

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Answer:

11. ASA

12. SSS

13. SAS

14. SAS

15. Not congruent

16. SAS

Step-by-step explanation:

11.

12, Oppositely congruent

13. indirectly congruent

14. Indirectly congruent

15.

16. Directly congruent

The value of A and B are -10x and -4 respectively.

Division involves gathering a number of expressions into smaller forms.

The quotient is represented by the expression

Quotient= Dividend /Divisor

We have the value of the Dividend from the division table given in the question

Dividend of the Operation= A

Divisor of the Operation=2

Quotient of the Operation= -5x

As we know Quotient= Dividend /Divisor

So, -5x= A/2

multiplying 2 on both sides

⇒(-5x)2= A/2*2

⇒ A= -10x

Similarly, We have the value of the Dividend from the division table given in the question

Dividend of the Operation= B

Divisor of the Operation=2

Quotient of the Operation = -2

As we know Quotient= Dividend /Divisor

So, -2= B/2

multiplying 2 on both sides

⇒(-2)2= AB2*2

⇒ B= -4

Therefore the value of A and B are -10x and -4 respectively.

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The value of x is 8 mm.

Similar Figures are the figures in which all the sides of one figure are in same ratio with the other side.

Two figures are given which are similar to each other .

When two figures are in ratio a:b then Area will be in ratio a² : b²

Area of figure 1 = 22 sq.mm

Area of figure 2 = 352 sq.mm

A1 :A2 = 22 /352

a² : b² = 1 / 16

a:b = 1/4

Therefore the bigger figure is 4 times the small figure .

The 4mm side will be 16 mm in the larger figure

The difference will be the value of x

x = 24 -16 = 8 mm

Therefore the value of x is 8 mm.

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Answer:

x = 8

Step-by-step explanation:

Similar shapes are dilations of each other using a scale factor.

As the area of both figures has been given, the scale factor that maps the green figure to the yellow figure can be calculated:

[tex]\implies \textsf{Scale factor (area)} = \dfrac{\textsf{area of yellow figure}}{\textsf{area of green figure}} = \sf \dfrac{352}{22} = 16[/tex]

Area is measured in square units. Therefore, to find the scale factor for length, square root the scale factor for area:

[tex]\begin{aligned}\implies \textsf{Scale factor (length)} & = \sf \sqrt{\textsf{Scale factor for area}} \\ & =\sf \sqrt{16}\\ & = \sf 4 \end{aligned}[/tex]

Use the found scale factor for length to find the perimeter of the yellow figure:

[tex]\begin{aligned} \sf \implies \textsf{Perimeter (yellow fig)} & = \sf \textsf{perimeter (green fig)} \times scale\:factor \\ & = \sf 26 \times 4 \\ & = \sf 104\: mm \end{aligned}[/tex]

Find the missing horizontal lengths of the yellow figure by using the corresponding length of the green figure:

⇒ 4 mm × 4 = 16 mm

⇒ 24 mm - 16 mm = 8 mm

Let the height of the yellow figure be y.

⇒ 24 + 8 + 16 + y + x + (y - x) = perimeter

⇒ 48 + 2y = 104

⇒ 2y = 56

⇒ y = 28 mm

Divide the yellow figure into two rectangles (see attachment).

To find x, sum the areas of the rectangles and equate to 352:

⇒ 224 + 16x = total area

⇒ 224 + 16x = 352

⇒ 16x = 128

⇒ x = 8

Step-by-step explanation:

the area of a parallelogram is

baseline × height

which is in our case

9 × 7 = 63 m²

Answer:

ayo give the other guy brainilest

Step-by-step explanation:

The decay factor of the equation given in the task content is; 0.87.

According to the task content; the equation given is; y = 453(0.87)^4.

By comparison with exponential decay functions, it follows that the decay factor from the equation given is; 0.87 which insinuates that the factor of change after each decay is 0.87.

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Answer:

The area of the shape is 68 cm².

Step-by-step explanation:

We have to calculate the areas of the rectangle and triangle separately, and then add them together.

• Area of rectangle = length × breadth

                                = 30 cm × 2 cm

                                = 60 cm²

• Area of triangle = [tex]\frac{1}{2}[/tex] × base × perpendicular height

                             = [tex]\frac{1}{2}[/tex] × 2 cm × 8 cm

                             = 8 cm²

Adding the areas together = 60 cm² + 8 cm²

                                             = 68 cm²

The special angles pairs that are formed are identified below.

Special angle pairs are formed by a transversal and the parallel lines it cuts across.

Based on the image given, the special angle pairs are given as follows:

Corresponding angles:

∠1 and ∠5

∠4 and ∠8

∠2 and ∠6

∠3 and ∠7

Alternate interior angles:

∠3 and ∠5

∠4 and ∠6

Consecutive interior angles:

∠4 and ∠5

∠3 and ∠6

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The value of x in the triangle will be 30°

Let the base angles be x. The vertex angle will be:

= 2(x + x) = 4x

Since the sum of the angles in a triangle is 180°, therefore, 4x + x + x = 180°

6x = 180°

x = 180°/6

x = 30°

Therefore, the angles are 30°, 30°, and 120°. This makes up 180°.

Complete question:

In an isosceles triangle, the vertex angle is twice the sum of the base angles. Fund the angles of the triangle.

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Considering the given exponential function, we have that:

An exponential function is modeled by:

[tex]y = ab^x[/tex].

In which:

When x = 0, y = 1, hence the initial value is of 1. When x changes by 1, y changes by 1/3, hence the base is of 1/3. Exponential functions have no restrictions, hence the domain is of all real values.

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Answer:

On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases into quadrant 2. It goes through (2, one-ninth), (1, one-third), (0, 1), and (negative 1, 3).

Analyze the graph of the exponential decay function.

The initial value is

✔ 1

The base, or rate of change, is

✔ 1/3

The domain is

✔ all real numbers

Step-by-step explanation:

Answer:

Step-by-step explanation:

The is a factor of -3 which makes a vertical stretch on the y-axis of the original f(x) equation making a new g(x) equation.

g(x) = a * f(b(x-h)) + k

a - vertical stretch/shrink y-axis

b - horizontal stretch/shirink x-axis

h - horizontal shift right/left

k - vertical shift up/down.

Desmos

Answer:

The probability that the number of people who show up will exceed the capacity of the plane is 0.002607722

The area of mathematics known as probability deals with numerical representations of the likelihood that an event will occur or that a statement is true. An event's probability is a number between 0 and 1, where, roughly speaking, 0 denotes the event's impossibility and 1 denotes certainty.

The probability that x people show up follows a binomial distribution,

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

Here, n represents the number of booked people and p represents the probability that a person shows up.

Then, the probability that x people show up is:

[tex]P(x)=\frac{78!}{x!(78-x)!}*0.90^{x}*(1-0.90)^{78-x}[/tex]

So, the probability that the number of people who show up will exceed the capacity of the plane is:

P(x>76) = P(77) + P(78)

Where P(77) and P(78) are calculated as follows:

[tex]P(77)=\frac{78!}{77!(78-77)!}* 0.90^{77}* (1-0.90)^{78-77} \\ =78*0.00029969*0.10\\ =0.0023375[/tex]

[tex]P(78)=\frac{78!}{78!(78-78)!}* 0.90^{78}* (1-0.90)^{78-78} \\ =1*0.00026972*1\\ =0.00026972[/tex]

Finally, P(x>76) is:

[tex]P(x > 76) = 0.0023375 + 0.00026972\\=0.002607722[/tex]

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Answer:

`1) 2.5cm, 2)1.25inches 3)2.25inches 4)1.25inches

Step-by-step explanation:

1) This is a centimeter rule, one small line represents 0.1cm and 1 long line represent 1cm, the line in between two long lines represent 0.5 cm

As you can see from A to B, is 2 cm + 0.5cm = 2.5cm.

2)This is found on the other side of rulers, inches.

1 small line represent 0.25 inches, line in between 2 long lines is 0.5 inches.

From S to T, is 1 inch + 0.25 inches = 1.25inches

3)Similar to Question 2.

Length of Pencil= 2 inches + 0.25 inches = 2.25 inches

4)This is a more detailed inch rule.

But length of heart shape = 1 inch + 0.25 inches = 1.25 inches

The number of combinations which are possible according to the specifications is; 119,700.

It follows from the task content that the 5ember teams to be formed must contain 3 girls and 2 boys.

On this note, it follows that the combinations possible are as follows

20C(3) × 15C2 = 1140 × 105 = 119,700.

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Answer:

[tex]f(x) = (x-12)(x-1)[/tex]

Step-by-step explanation:

So when you have a quadratic equation it can usually be expressed as: [tex]a(x-b)(x-c)[/tex] where b and x are the roots of the equation. Since it's just asking for a function whose zeroes are 12 and 1, I'm assuming it doesn't care about the stretch/compression (value of a determines that), so I'll just assume a=1. if 12 and 1 are zeroes, that means it makes one of the factors equal to 0 because is x-b = 0 then no matter the value of x-c the value of y is 0 because 0 * (x-c) will be 0. So the equation becomes [tex]f(x) = (x-12)(x-1)[/tex] plugging in 12 or 1 as x will make one of the factors 0 thus making f(x) equal to 0

Answer:

See proof below

Step-by-step explanation:

Given [tex]\triangle ADE \cong \triangle CBD[/tex]

Then, [tex]\angle AED \cong \angle CFB[/tex] because corresponding parts of congruent triangles are congruent.

Since [tex]\angle AED \text{ and } \angle DEB \text{ form a linear pair}[/tex], by the linear pair postulate, [tex]\angle AED \text{ and } \angle DEB \text{ are supplementary}[/tex].

Similarly, [tex]\angle CFB \text{ and } \angle BFD \text{ form a linear pair}[/tex], so by the linear pair postulate,  [tex]\angle CFB \text{ and } \angle BFD \text{ are supplementary}[/tex].

By the Congruent supplements theorem, since [tex]\angle AED \text{ and } \angle DEB \text{ are supplementary}[/tex], [tex]\angle CFB \text{ and } \angle BFD \text{ are supplementary}[/tex], and [tex]\angle AED \cong \angle CFB[/tex] , then [tex]\angle DEB \cong \angle BFD[/tex]. (note, this is one pair of opposite angles inside quadrilateral DFBE)

Recalling that [tex]\overline {ED} \text{ bisects } \angle ADC, \text{ and } \overline {FB} \text{ bisects } \angle ABC[/tex], then, [tex]\angle ADE \cong \angle CDE[/tex], and [tex]\angle CBF \cong \angle FBA[/tex] by definition of angle bisector.

Note that [tex]\angle ADE \cong \angle CBF[/tex] because corresponding parts of congruent triangles are congruent.

Also, note that [tex]\angle EDF \cong \angle EDC[/tex] and [tex]\angle FBA \cong \angle FBE[/tex] because B, E, A are colinear, and D, F, C are colinear.

So, by the transitive property of angle congruence, [tex]\angle EDF \cong \angle FBE[/tex]  (This is the other pair of opposite angles inside quadrilateral DFBE)

So, since both pairs of opposite angles are congruent, quadrilateral DFBE is a parallelogram, by a theorem about quadrilateral properties (your book may or may not have a name for it.  It may just have a number).

Answer:

         (8b − 1)b = 8b*b - 1*b

                        = 8b2 - b

Step-by-step explanation:

Hence the expression (8b − 1)b=8b2 - b.

This shows that if s is doubled and h tripled, then V multiplied by 12

Given the formula below expressed as:

V = s²h

If s is doubled and h tripled, the new volume will ne:

V1 = (2s)² * 3h

V2 = 4s² * 3h

V2 = 12s²h

V2 = 12V

This shows that if s is doubled and h tripled, then V multiplied by 12

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There is a significant value in the difference in the value of the car after nine years

The table of values is given as:

Years     1     2     3

Car 1 17,390 16,346.60 15,365.80

Car 2 17,500 16,500 15,500

The values of car 1 have a common ratio of 0.94; this is calculated by dividing the terms of the function

The common ratio implies that the function of car 1 is a geometric function.

The values of car 2 have a common difference of -1000.

The common difference implies that the function of car 2 is a linear function.

Car 1

In (a), we have:

Common ratio, r = 0.94

The initial value is given as:

a = 18,500

So, the function of car 1 is

f(x) = 18500 * 0.94^x

Car 2

In (a), we have:

Common difference, d = -1000

The initial value is given as:

a = 18,500

So, the function of car 2 is

f(x) = 18500 - 1000x

This means that

x = 9

So, we have:

Car 1

f(9) = 18500 * 0.94^9

f(9) = 10600

Car 2

f(9) = 18500 - 1000 *9

f(9) = 9500

10600 is greater than 9500

Hence, there is a significant value in the difference in the value of the car after nine years

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Answer:

1)

Let the number of guests = x

∴ Pin hall total charge = 500 + 15x

Bloom place total charge = 350 + 18x

2)

The charges are equal.

∴  500 + 15x = 350 + 18x

⇒ 150 = 3x

⇒ 50 = x

∴ x = 50

This means that the charges for both halls will be equal when there are 50 guests.

[tex]C_{ij}[/tex] is the entry of [tex]C[/tex] in row [tex]i[/tex] and column [tex]j[/tex]. [tex]C[/tex] has 3 rows and 2 columns, so

[tex]C_{11} = \dfrac{|-2\times1+3\times1|}2 = \dfrac{|-2+3|}2 = \dfrac{|-1|}2 = \dfrac12[/tex]

[tex]C_{12} = \dfrac{|-2\times1+3\times2|}2 = \dfrac{|-2+6|}2 = \dfrac{|4|}2 = 2[/tex]

[tex]C_{21} = \dfrac{|-2\times2+3\times1|}2 = \dfrac{|-4+3|}2 = \dfrac{|-1|}2 = \dfrac12[/tex]

[tex]C_{22} = \dfrac{|-2\times2+3\times2|}2 = \dfrac{|-4+6|}2 = \dfrac{|2|}2 = 1[/tex]

[tex]C_{31} = \dfrac{|-2\times3+3\times1|}2 = \dfrac{|-6+3|}2 = \dfrac{|-3|}2 = \dfrac32[/tex]

[tex]C_{32} = \dfrac{|-2\times3+3\times2|}2 = \dfrac{|-6+6|}2 = \dfrac{|0|}2 = 0[/tex]

Then the matrix is

[tex]C = \begin{bmatrix}\dfrac12 & 2 \\\\ \dfrac12 & 1 \\\\ \dfrac32 & 0\end{bmatrix} = \dfrac12 \begin{bmatrix}1&4\\1&2\\3&0\end{bmatrix}[/tex]

By summing the areas of minor rectangles and triangles, the area of the irregular polygon whose vertices are (- 3, 4), (2, 4), (4, 2), (2, - 3) and (- 3, - 3) is equal to 42 square units.

In this problem we have to determine the area of a irregular polygon, whose calculation is defined as a sum of the areas of triangles and rectangles. To determine the combination of squares and triangles, we need first to plot the vertices and construct the figure on a Cartesian plane.

The area of the irregular figure is:

A = (5) · (7) + 0.5 · 2² + 0.5 · (2) · (5)

A = 35 + 2 + 5

A = 42

By summing the areas of minor rectangles and triangles, the area of the irregular polygon whose vertices are (- 3, 4), (2, 4), (4, 2), (2, - 3) and (- 3, - 3) is equal to 42 square units.

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Answer: [tex][-3,7][/tex]

Step-by-step explanation:

The range of a function is the set of y-values.

The area of the shaded region = π(4²) - (4√2)² = 50.27 - 32

The area of the shaded region, to one decimal place is: 18.3 m²

The area of the shaded region = area of the circle - area of the square = πr² - s².

Radius (r) of the circle = 1/2(length of diagonal of the square)

Using Pythagorean theorem, we have:

Diagonal = √[(4√2²) + (4√2²)]

Diagonal = √(32 + 32]

Diagonal = 8 m

Radius (r) = 8/2 = 4

s = 4√2

The area of the shaded region = π(4²) - (4√2)² = 50.27 - 32

The area of the shaded region = 18.3 m²

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Answer:

The zeroes are 2, -2, -6, 5.

Step-by-step explanation:

If 2 of the zeroes are 2 and -2 then (x - 2)(x + 2) is factor of the polynomial.

We can now do long division.

Note (x - 2)(x + 2) = x^2 - 4, so we have:

x^2 - 4 ( x^4 + x^3 - 34x^2 - 4x + 120 ( x^2 +   x  - 30 <--- Quotient

             x^4            - 4x^4

                      x^3 -  30x^2 - 4x

                      x^3                -4x

                                  -30x^2   + 120  

                                  -30x^2   + 120

x^2 +   x  - 30 = 0        

(x + 6)(x - 5) = 0

x = -6, 5.              

Answer:

5 and -6

Step-by-step explanation:

So when a polynomial has zeroes at a, then one of the factors (x-a). This is because in factored form, a polynomial can be written using it's factors as such: [tex]a(x\pm b)(x\pm c)(c \pm d)...[/tex] and so on, with the number of factors depending on the degree. The sign is plus or minus because it can be either, it will depend on the sign of the zero. The reason the zeroes are factors, is because when one of the factors is zero it makes the entire thing zero because multiplying them out will always get 0, if you're multiplying by a zero. Anyways, if you have a factor, you can divide, by those factors. So let's do that.

So you have the two factors (x+2)(x-2). So let's divide by each factor. In this example I'll be using long division. So essentially what long division, is finding how many times the leading coefficient of the dividend (x^4+x^3-34x^2-4x+120) can be divided by the leading coefficient of the divisor ((x-2)(x+2)). And then whatever you get, that will be part of the quotient, then take whatever you got, and multiply it by the divisor and then subtract that product from the dividend and then continue this process until the dividend is equal to 0 or you can't divide any further./

Original equation:

[tex](x+2)\div (x^4+x^3-34x^2-4x+120)[/tex]

Divide the leading coefficients:

[tex]\frac{x^4}{x}=x^{4-1} = x^3[/tex]

The x^3 will be part of the quotient, now multiply this by the divisor

[tex]x^3(x+2) = x^{3+1} + 2x^3 = x^4 + 2x^3[/tex]

Subtract the product from the dividend

[tex](x^4+x^3-34x^2-4x+120) - (x^4+2x^3) = x^4+x^3-34x^2-4x+120 - x^4 - 2x^3[/tex]

Group like terms:

[tex](x^4 - x^4) + (x^3 - 2x^3) -34x^2-4x+120[/tex]

Simplify:

[tex]-x^3 -34x^2-4x+120[/tex]

Divide leading coefficients:

[tex]\frac{-x^3}{x} = -x^{3-1} = -x^2[/tex]

The -x^2 will be part of the quotient, now multiply this by the divisor:

[tex]-x^2(x+2) = -x^{2+1} -2x^2 = -x^3-2x^2[/tex]

Subtract this from the dividend:

[tex](-x^3 -34x^2-4x+120) - (-x^3 - 2x^2) = -x^3 -34x^2-4x+120 + x^3 + 2x^2[/tex]

Group like terms:

[tex](-x^3 + x^3) + (-34x^2 + 2x^2)-4x+120[/tex]

Simplify:

[tex]-32x^2 -4x -120[/tex]

Divide the leading coefficients:

[tex]-\frac{32x^2}{x}=-32x^{2-1} = -32x[/tex]

This will be part of the quotient, now multiply this by the divisor:

[tex]-32x(x+2) = -32x^2 - 64x[/tex]

Subtract this from the dividend

[tex](-32x^2 -4x -120) - (-32x^2-64x) = -32x^2 -4x -120 + 32x^2 + 64x[/tex]

Group like terms:

[tex](-32x^2+32x^2) + (-4x+64x) -120[/tex]

Simplify:

[tex]60x+120[/tex]

Divide the leading coefficients:

[tex]\frac{60x}{x} = 60[/tex]

This will be part of the quotient, multiply it by the divisor:

[tex]60(x+2) = 60x+ 120[/tex]

Subtract this from the dividend:

[tex](60x+120)-( 60x+120) = 0\\[/tex]

Since this comes out perfectly to zero there is no remainder, and this is expected since it's a factor. You now take all the quotients from dividing the leading terms and you get [tex]x^3-x^2-32x+60[/tex]. To keep the original equation you simply multiply it, and you get: [tex](x+2)(x^3-x^2-32x+60)[/tex]. But what we really care about is the x^3-x^2-32x+60. We now divide it by the other factor x-2 using the same process

Original equation:

[tex]x^3-x^2-32x+60[/tex]

Divide leading terms:

[tex]\frac{x^3}{x} = x^2[/tex]

This will be part of the quotient, multiply it by the divisor

[tex]x^2(x-2) = x^3-2x^2[/tex]

Subtract this from the dividend

[tex](x^3-x^2-32x+60) - (x^3 - 2x^2) = x^2-32x+60[/tex]

Dividing leading coefficients:

[tex]\frac{x^2}{x} = x[/tex]

This will be part of the quotient, multiply it by the divisor:

[tex]x(x-2) = x^2-2x[/tex]

Subtract this from the dividend:

[tex](x^2-32x+60) - (x^2-2x) = -30x+60[/tex]

Divide the leading terms:

[tex]\frac{-30x}{x} = -30[/tex]

This will be part of the quotient, now multiply it by the divisor:

[tex]-30(x-2) = -30x+60[/tex]

Subtract it from the dividend:

[tex](-30x + 60) - (-30x + 60) = 0[/tex]

Now take all the quotients of the leading terms and you get:

[tex]x^2+x-30[/tex]

Now we can factor this quadratic to get the equation in completely factored form: We're looking for factors that multiply to -30 and add to 1. So we really need to look for factors that have a difference of 1, Or in other words |a-b| = 1. The sign can be determined after we fine the two factors. The factors 5 and 6 have a difference of 1, and since it needs to add to 1, the 5 has to be negative, and the 6 has to be positive.

So we get the factors: [tex](x-5)(x+6)[/tex]. Setting each of them equal to zero gets you:

x-5 = 0

x = 5

x + 6 = 0

x = -6

So the other two zeroes are 5 and -6