If,0 < 3x - 2 < 1, what is one possible value for x?​

Answer:

Step-by-step explanation:

5x-9 = 2(x+23) or 5x-9 = 2x+46

Let' x be the "number."

"Five times a number decreased by nine" means: 5x - 9

"equal to twice the number increased by 23" means:  = 2(x+23)

Put them together:  5x-9 = 2(x+23)

5x-9 = 2(x+23)

5x-9 = 2x+46

[The formatting of the possible answers does not include parentheses nor algebraic symbols, making selection of a correct answer difficult. ]

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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We gujra weeeeeeee

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Answer:

Step-by-step explanation:

at first decreased by 60% now increased by 80%

100 - 60% = 40

40 + 80% =

40 + 40 :100 * 80 =

40 + 32 =

72

Answer: 1344

Step-by-step explanation:

Using the Pythagorean theorem, if we consider the unknown leg of the base, we get that it has a length of 12.

So, the area of the base is 96.

Multiplying this by the height, we get the volume is 1344 cubic inches.

Answer: B

Step-by-step explanation:

Given:
36=<1.02D-31.2<=41

Solve the compound inequality for D:

36+31.2<=1.02D<=41+31.2

67.2<=1.02D<=72.2

65.9<=D<=70.8

Therefore, option (b) is correct answer

Answer:

The linear equation for the line with an x - intercept at (-2,0) and y-intercept at (0,-3) is found as [tex]$y=-\frac{3}{2} x-3$[/tex].

Step-by-step explanation:

A condition is given that a line has an x- intercept at (-2,0) and y - intercept at (0,-3).

It is asked to find a linear equation satisfying the given condition.

Step 1 of 2

Determine the slope of the line.

The points of the intercepts of the line are given as (-2,0) and (0,-3). Next, the formula for the slope is given as,

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$[/tex]

Substitute -3&0 for [tex]$y_{2}$[/tex] and [tex]$y_{1}$[/tex] respectively, and 0&-2 for [tex]$x_{2}$[/tex] and [tex]$x_{1}$[/tex] respectively in the above formula. Then simplify to get the slope as follows,

[tex]$$\begin{aligned}m &=\frac{-3-0}{0-(-2)} \\m &=\frac{-3}{2} \\m &=-\frac{3}{2}\end{aligned}$$[/tex]

Step 2 of 2

Write the equation in the slope-intercept form.

The slope-intercept form of a line is given as follows,

y=mx+b

The coordinates at the y- intercept is (0,-3). Now, as the y- coordinate is -3, so b=-3.

So, substitute -3 for b and [tex]$-\frac{3}{2}$[/tex] for m in the equation y=mx+b, and simplify to get the equation as follows,

[tex]$$\begin{aligned}&y=-\frac{3}{2} x+(-3) \\&y=-\frac{3}{2} x-3\end{aligned}$$[/tex]

This is the required linear equation.

The method that Chris can use to make sure the segments in the pattern are all parallel is to treat AI as a single line that acts as a transversal for the lines that need to be parallel.

Within the set of lines which need to be parallel, the acute angles must be congruent.

Both angles CDE and EDF need to be 180° minus twice the measure of one of the congruent acute angles.

The pair of the acute angles previously within each of the two sets of parallel lines are corresponding.

The fact that the transversals go through endpoints, there are no alternate angles, interior or exterior because alternate angles are on opposite sides of the transversal.

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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:

Graph of exponential rising up to the right, through the point (0, 2).

Step-by-step explanation:

I think you mean  f(x) = 3^x + 1.

When x = 0, 3^x = 1, so the graph of f(x) = 3^x will rise up to the right and pass through point (0. 1)

The + 1 translates the graph up 1 unit so

The graph of f(x) = 3^x + 1 passes through the point (0, 2).

Answer:

1200

Step-by-step explanation:

Ok since the oldest was given 2/5, the middle was given 1/2, and the youngest was given the rest, that means the fraction that the youngest was given has to make 2/5 + 1/2 + ? = 1. So we have the equation: [tex]\frac{2}{5} + \frac{1}{2} + x = 1[/tex]. Now to add the fractions simply multiply 2/5 by 2/2 and multiply 1/2 by 5/5 To get the same denominator: [tex]\frac{4}{10} + \frac{5}{10}+x=\frac{9}{10}+x[/tex]. This means that x is 1/10 , because in adding 1/10 you get the whole 10/10. So 120 is one tenth of the entire prize. This means to get the entire prize you simply multiply by 10. This gives you the equation 120 * 10 = 1200

i. 20 people

ii. 3 people

iii. 26 people

We have that;

3x + 5 people likes apples

3x + 5 people likes banana

2x + 15 people likes at least one of the fruits

The total number of people = 55

Let's find the value of x

3x + 5 + 2x + 15 = 55

5x + 20 = 55

5x = 55 - 20

5x = 35

x = 35/5

x = 7

i. People that likes banana only =  3x + 5 = 3 × 5 + 5 = 15 + 5 = 20 people

ii. None of the fruits = 55 - 2x + 15 = 55 - 3x + 5 - 3x + 5 = 55 - 26 +26

= 55 - 52

= 3 people

iii. At most one fruit = 55 - 2x + 15 = 55 - 2(7) + 15 = 55 - 29 = 26 people

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A chord is a part of a circle that joins two points on its circumference but does not pass through the center of the circle. From the given diagram in the question, GH = HJ. Thus the measure of chord HJ is option C. 8.6 units.

A chord is one of the parts of a circle which does not pass through its center but joins two points on its circumference. So that it is not the same as the radius of the circle.

In the given question, it has been stated that: GH and HJ are congruent.

So that by the congruency property of two lines, the length of chod GH is the same as the length of chord HJ.

Therefore, the measure of chord Hj is 8.6 units. This implies option C.

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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:

negative y axis

Step-by-step explanation:

6 radians is between 3pi/2 and 2pi radians, so it lies in tje fourth quadrant.

So, a terminal side is the negative y axis.

The ordered pair which makes both inequalities true is: D. (3, 0).

In Mathematics, an inequality can be used to show the relationship between two (2) or more integers and variables in an equation.

In order to determine ordered pair which makes both inequalities true, we would substitute the points into the inequalities as follows:

At (0, 0), we have:

y > -2x + 3  

0 > -2(0) + 3

0 > 3 (false).

y < x – 2

0 < 0 - 2

0 < -2 (false)

At (0, -1), we have:

y > -2x + 3  

-1 > -2(0) + 3

-1 > 3 (false).

y < x – 2

-1 < 0 - 2

-1 < -2 (false)

At (1, 1), we have:

y > -2x + 3  

1 > -2(1) + 3

1 > -1 (true).

y < x – 2

1 < 1 - 2

1 < -1 (false)

At (3, 0), we have:

y > -2x + 3  

0 > -2(3) + 3

0 > -3 (true).

y < x – 2

0 < 3 - 2

0 < 1 (true).

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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

Using an exponential function, we have that:

a) The equation is: [tex]A(t) = 600000(1.058)^{\frac{t}{2}}[/tex].

b) The house will have a value of $1 million during the year of 2037.

c) The expected value on April 1, 2020, is of $621,520.

An increasing exponential function is modeled by:

[tex]A(t) = A(0)(1 + r)^t[/tex]

In which:

In this problem, the parameters are given as follows:

A(0) = 600000, r = 0.058 each 2 years.

Hence the exponential function is:

[tex]A(t) = 600000(1 + 0.058)^{\frac{t}{2}}[/tex]

[tex]A(t) = 600000(1.058)^{\frac{t}{2}}[/tex]

Item b:

It is the year 2019 + t, for which A(t) = 1000000, hence:

[tex]A(t) = 600000(1.058)^{\frac{t}{2}}[/tex]

[tex]1000000 = 600000(1.058)^{0.5t}[/tex]

[tex](1.058)^{0.5t} = 1.6667[/tex]

[tex]\log{(1.058)^{0.5t}} = \log{1.6667}[/tex]

[tex]0.5t\log{1.058} = \log{1.6667}[/tex]

[tex]t = \frac{\log{1.6667}}{0.5\log{1.058}}[/tex]

[tex]t = 18.12[/tex]

The house will have a value of $1 million during the year of 2037.

Item c:

April 1, 2020 is one year and 3 months = 1.25 years after January 1, 2019, hence the value of the home will be given by:

[tex]A(t) = 600000(1.058)^{\frac{t}{2}}[/tex]

[tex]A(1.25) = 600000(1.058)^{\frac{1.25}{2}}[/tex]

A(1.25) = $621,520.

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Answer:

Given :- ABCD is a square.

To proof :- AC = BD and AC ⊥ BD

Proof :- In △ ADB and △ BCA

AD = BC [ Sides of a square are equal ]

∠BAD = ∠ABC [ 90° each ]

AB = BA [ Common side ]

△ADB ≅ △BCA [ SAS congruency rule ]

⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]

In △AOB and △AOD

OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]

AB = AD [ Sides of a square are equal ]

AO = AO [ Common side ]

△AOB ≅ △ AOD [ SSS congruency rule ]

⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]

∠AOB + ∠AOD = 180° [ Linear pair ]

∠ AOB = ∠AOD = 90°

⇒ AO ⊥ BD

⇒ AC ⊥ BD

Hence proved, AC = BD and AC ⊥BD

The initial price in this scenario is > sh 276000 and was therefore sold at a loss.

This is a situation in which the selling price of goods or services is lower than the cost price.

We were told that the selling price is sh 276000 which is less than the initial(cost) price hence he ran at a loss during the business transaction.

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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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[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]

Answer:

x ≈ 8.0

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos37° =  =  ( multiply both sides by 10 )

10 × cos37° = x , thus

x ≈ 8.0 ( to the nearest tenth )

does this help? :D

Answer:

  30.70 ft

Step-by-step explanation:

For shadow problems, we make the assumption that an object's height and its shadow length are proportional.

The proportion for the two objects is ...

  height/shadow = pole/(22 ft) = (6 ft)/(4.3 ft)

Multiplying by 22 ft, we find the pole height to be ...

  pole = (22 ft)(6 ft)/4.3) ≈ 30.70 ft

The flagpole is about 30.70 feet tall.

Answer:

[tex] \frac{ - 0.5}{ - 0.5} = 1 \\ [/tex]

1 is a whole number, an integer, and rational.

Step-by-step explanation:

The complete similarity transformation is: triangle abc was dilated by a scale factor of [2], reflected across the [x-axis], and moved through the translation [0].​

From the figure, we have:

AB = 2 units

DE = 4 units

So, the scale factor (k) is:

k = DE/AB

k = 4/2

k = 2

Also, both shapes are on either sides of the x-axis, and they are equidistant from the x-axis.

This means that the triangle is reflected across the x-axis with no translation.

Hence, the complete similarity transformation is: abc was dilated by a scale factor of [2], reflected across the [x-axis], and moved through the translation [0].​

So, the scale factor is 2

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The equation that represents the line that passes through the points is [tex]y-1 = \frac{5}{4}(x-1)[/tex]. The correct option is D. [tex]y-1 = \frac{5}{4}(x-1)[/tex]

From the question, we are to determine the equation of the line that passes through the given points

Using the formula

[tex]\frac{y-y_{1} }{x-x_{1}} =\frac{y_{2}-y_{1} }{x_{2}-x_{1}}[/tex]

From the give information,

[tex]x_{1} = 1 \\y_{1} = 1\\{x_{2} =5[/tex]

[tex]y_{2} = 6\\[/tex]

Putting the parameters into the formula,

[tex]\frac{y-1 }{x-1} =\frac{6-1}{5-1}[/tex]

[tex]\frac{y-1 }{x-1} =\frac{5}{4}[/tex]

∴ [tex]y-1 = \frac{5}{4}(x-1)[/tex]

Hence, the equation that represents the line that passes through the points is [tex]y-1 = \frac{5}{4}(x-1)[/tex]. The correct option is D. [tex]y-1 = \frac{5}{4}(x-1)[/tex]

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