Graph the inequality. -3 < y - 7x < 15

The expressions are not equivalent, even though Ella got the same result when she substituted zero for f. Then the correct option is C.

The equivalent is the expressions that are in different forms but are equal to the same value.

Ella completed the following work to test the equivalence of two expressions.

First expression,                Second expression

⇒ 2f + 2.6                           ⇒ 3f + 2.6

⇒ 2(0) + 2.6                       ⇒ 3 (0) + 2.6

⇒ 0 + 2.6                           ⇒ 0 + 2.6.

⇒ 2.6                                 ⇒ 2.6

The expressions are not equivalent, even though Ella got the same result when she substituted zero for f.

Then the correct option is C.

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The amount of Type A coffee that was used is 40 pounds.

From the given information;

Type A coffee costs Miguel $5.95 per pound

Type B coffee costs $4.65 per pound.

For this month, the equation can be expressed as:

A + 3A

Thus;

5.95A + 3(4.65)A = 796.00

5.95A +13.95A = 796.00

19.9A = 796.00

A = 796.00/19.9

A = 40

Therefore, we can conclude that the amount of pounds of type A coffee that was used is 40 pounds.

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By ASA postulate it can be say that the triangle ABC and triangle MRQ are congruent

The sum of all three angles(interior) of a triangle is 180 degrees, and the exterior angle of a triangle measures the same as the sum of its two opposite interior angles.

Using Angle Sum Property, in ΔMRQ

<M + <R + <Q= 180

<M + 42 + 85 = 180

<M = 180 -42

<M= 53°

Now, In ΔMRQ and ΔABC  

<A= <R ,

<B = <M

AB = MR

Hence, By ASA criteria ΔMRQ ≅ ΔABC  

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Answer:

Step-by-step explanation:

m represents the meters.

Answer:

They have none left over and they can make 43 dishes

Step-by-step explanation:

The original price is £75

Let the original price be x.

The reduced price is given as:

Reduced =60

So, we have:

Reduced = Original price * (1 - discount)

This gives

x * (1- 20%) = 60

Evaluate the difference

x * 0.8 = 60

Divide by 0.8

x = 75

Hence, the original price is £75

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The bat will not fit in the cube shaped box

The side length of the cube is given as:

x = 17

The diagonal of the cube is calculated as

[tex]d = x\sqrt 3[/tex]

So, we have:

[tex]d = 17\sqrt 3[/tex]

Evaluate

d = 29.4

29.4 is less than the length of the bat (30 inches)

This means that the bat will not fit in the cube shaped box

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The congruence theorem that can be used is: B. ASA

If we have two triangles that have two pairs of corresponding congruent angles (e.g. ∠LGH ≅ ∠HKJ and ∠LHG ≅ ∠KHJ), and a pair of corresponding congruent sides (e.g. GH ≅ HK), the triangles are said to be congruent triangles by the ASA congruence theorem.

Therefore, triangles GHL and KHL in the image given are congruent triangles by the  ASA congruence theorem.

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Answer: Option 2 or B. ASA

Step-by-step explanation: Trust me!

If you are working on Edge the answer is correct. The proof is down below. I hope someone finds this helpful. Good luck everyone!  :)

The length of the run is at most 15 yards

The given parameters are:

Width = 5 yards

Height = x

Perimeter = Not more than 40 yards

The perimeter is calculated as:

P  = 2 *(Width + Height)

So, we have:

2 *(5 + x) ≤40

Divide by 2

5 + x ≤ 20

Subtract 5 from both sides

x ≤ 15

This implies that the length of the run is at most 15 yards

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Answer: C

Step-by-step explanation:

12in is 1 ft

652/12 is 54r4

so atlease 55

Answer:

13.8

Step-by-step explanation:

because they're similar, we know that

zy ÷ wz has the same ratio as xy ÷ wv

12 / 15 = 11 / wv

cross multiply

12wv = 165

÷ 12     ÷12

wv = 13.75

13.75 rounded to nearest tenth is 13.8

Option B is correct. The statement that is true about the values we have been given here is that They are the only values that make both equations true.

We have been given two equations

The first equation is

4y-3x=16

We have to plug in the values (-5, 0.25) in this equation

4(0.25) -3(-5) =16

Then we have

1 + 15 = 16

Hence 16 = 16 This satisfies the equation.

In the second equation we have to follow the same step

-x-8y=3

Such that

-(-5)-8*0.25 = 3

5-2 = 3

3 = 3 This also satisfies the equation.

Therefore the correct answer option is B

a. They satisfy equation 8 but not equation A.

B. They are the only values that make both equations true.

C. They show that the equations represent the same line.

D. They satisfy equation A but not equation B.

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The relative frequency of the class interval 65-75 is .20.

The word 'relative' is used to indicate that an event is being considered in relation or in proportion to something else.

Frequency is a way to measure how often a particular event occurs. Relative frequency, on the other hand, is a way to measure how often a particular event occurs against total occurrences.

Given that: 50 participants scored between 65 and 75 on a test.

Hence,

The relative frequency of the class interval 65-75 = 75 - 65 = 20

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Answer:

200

Step-by-step explanation:

Solution,

Given,

n(A)=100

n(B)=120

A⊂B

n(A∩B)=20

n(A∪B)=n(A)+n(B)-n(A∩B)

n(A∪B)=100+120-20

n(A∪B)=200

n(A union B) is equal to the number of elements in set B, which is 120.

If A is a subset of B, it means that every element in set A is also an element of set B.

In this case, n(A) represents the number of elements in set A, which is 100, and n(B) represents the number of elements in set B, which is 120.

To find the value of n(A union B) (the number of elements in the union of sets A and B), we need to consider that the union of two sets includes all the elements that are in either set A or set B, or both.

Since A is a subset of B, all the elements in set A are already included in set B. Therefore, the union of sets A and B will be equal to set B itself.

Thus, n(A union B) is equal to the number of elements in set B, which is 120.

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The constant of proportionality in centimeters per minute is: 5.5.

To find the constant of proportionality of the table shown in the image attached below, use one pair of points, (12, 66).

Constant of proportionality, k = y/x, therefore, the proportional constant in centimeters per minute would be:

k = 66/12 = 5.5 cm per min.

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Answer:

x = 80

Step-by-step explanation:

if ray RU bisects <QRS that means it divided the angle into two equal parts

then we can write the following equation to find the value of x:

<SRU = <QRU

50 = (x/2 + 10) subtract 10 from both sides

40 = x/2 multiply both sides with 2

80 = x

<QRS = 100 because it's the sum of <SRU and <QRU

Answer:

Step 3 is wrong and the answer should be 576

Step-by-step explanation:

Step 1 and step 2 are correct

Step 2: 4 × 12^2

You don't multiply 12 by 4, you first complete the exponent since, in PEMDAS, Exponent is before Multiplying.

Step 3 should be: 4 × (12×12)

= 4 × 144

= 576

Considering the volume of the given rectangular prism, the height is of 5.03 feet.

The volume of a rectangular prism of length l, width w and height h is given by:

V = lwh.

In this problem, we have that:

The volume is of 365 cubic feet, hence:

lwh = 365

(2h + 2)(h + 1)h = 365

(2h² + 4h + 2)h - 365 = 0

2h³ + 4h² + 2h - 365 = 0.

Using a cubic equation calculator, the height is of h = 5.03 feet.

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Answer:

x= 3

y=2

Step-by-step explanation:

Let's solve your equation step-by-step.

2=4x−10

Step 1: Flip the equation.

4x−10=2

Step 2: Add 10 to both sides.

4x−10+10=2+10

4x=12

Step 3: Divide both sides by 4.

4x/4=12/4

x=3

Answer:

since y=2 substitute y into y=4x-10

2=4x-10

2+10=4x

12=4x

Divide both sides by 4

12/4=4x/4

x=3

Therefore the answer is (2,3)

y= 4x-10

2= 4×3-10

2= 12-10

2=2

The area or the common pasture will be 12 acres.

Based on the information given, the formula to use will be:

ab/ac = 2/4

b/c = 1/2 .... i

Also, ab/bd = 2/6

a/d = 1/3 .... ii

Multiply both equation

1/2 × 1/3 = 1/6

Therefore, the area will be

= 6 × 2

= 12 acres

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The model (x+a)(x-a) will represents the factors of 4x²-9 as (2x+3)(2x-3).

The "completing the squares" method aims to construct a quadratic equation of the form where the x variable is entirely covered by a single squared term, which is the square of a linear expression in x, as the name suggests.

The quadratic equation can resemble this:

x²-a² = (x+a)(x-a)

Given equation;

⇒4x²-9

⇒(2x)²-3²

The equation can be modeled as;

(2x+3)(2x-3)

Hence the model (x+a)(x-b) will represents the factors of 4x²-9 as (2x+3)(2x-3).

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Answer:

If a function is continuous and one - to - one then it is either always increasing or always decreasing.

Step-by-step explanation:

An easy way to see this on a graph is to draw a horizontal line through the graph . If the line only cuts the curve once then the function is one - to - one.

Answer:

If a one-to-one function is not continuous, it is sometimes increasing or decreasing. If a one-one function is continuous, it is always increasing or always decreasing.

For example, if [tex]$f(x)=\left\{\begin{array}{lll}-x & \text { if } & |x| \geq 1 \\ x & \text { if } & |x| < 1\end{array}\right.$[/tex]

This function is one-to-one and it is not always increasing or decreasing.

Step-by-step explanation:

A statement that are one-to-one functions either always increasing or always decreasing is given.

It is required to explain the given condition.

To explain the given condition, consider a continuous and a not continuous function then explain about the one-to-one function.

If a one-to-one function is not continuous, it is sometimes increasing or decreasing.

For example, if [tex]$f(x)=\left\{\begin{array}{lll}-x & \text { if } & |x| \geq 1 \\ x & \text { if } & |x| < 1\end{array}\right.$[/tex]

This function is one-to-one and it is not always increasing or decreasing.

If a one-one function is continuous, it is always increasing or always decreasing.

For example, if f(x)=x-3. It is a continuous function and it reaches a minimum value at some values of x and it then keeps increasing.

There will be the same output value for different input values. It does not pass the horizontal line test and so it is not a one-to-one function.

In rectangles P and Q, where the area of rectangle Q is 10% less than the area of rectangle P, we have the ratio of x:y = 25:18.

The area of a rectangle is the product of its length and its width.

In the question, we are given two rectangles:-

Rectangle P:

length = 40 cm,

width = x cm,

Thus, the area = length * width = 40x.

Rectangle Q:

length = 25% more than the length of P = 40 + 25% of 40 = 40 + 0.25*40 = 40 + 10 = 50 cm.

width = y cm,

Thus, the area = length * width = 50y.

Now, we are given that, the area of rectangle Q is 10% less than the area of rectangle P.

Thus, we can write that,

Area of rectangle Q = Area of rectangle P - 10% of the area of rectangle P,

or, 50y = 40x - 10% of 40x,

or, 50y = 40x - 0.10*40x,

or, 50y = 40x - 4x,

or, 50y = 36x,

or, y = 36x/50.

We are asked to find the ratio of x:y = x/y.

Substituting y = 36x/50, we get,

x/y = x/(36x/50) = 50x/36x = 25/18.

Thus, the ratio x:y = 25:18.

Thus, in rectangles P and Q, where the area of rectangle Q is 10% less than the area of rectangle P, we have the ratio of x:y = 25:18.

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Answer:

  x ≈ -8.402, 8.502

Step-by-step explanation:

The usual methods of solving quadratic equations apply to this one. Any of them can be used: graphing, factoring, completing the square, quadratic formula.

Often, when the coefficients are non-integers and have no obvious relationship to each other, the most convenient method of solution is the quadratic formula. It tells you the solution to ax²+bx+c=0 is ...

  [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

Your equation has the coefficients, a = -0.07, b = 0.007, c = 5, so the solutions are ...

  [tex]x=\dfrac{-0.007\pm\sqrt{0.007^2-4(-0.07)(5)}}{2(-0.07)}=\dfrac{-0.007\pm\sqrt{1.400049}}{-0.14}\\\\x=0.05\pm\sqrt{\dfrac{200007}{2800}}=0.05\pm\sqrt{71.4310\overline{714285}}\\\\\boxed{x\approx\{-8.40169,\,8.50169\}}[/tex]

Answer:

[tex]t=8.50, -8.40\:\: \sf (2 \:d.p.)[/tex]

Step-by-step explanation:

Quadratic equations are in the form [tex]ax^2+bx+c=0[/tex],  where [tex]a\neq 0[/tex].

There are three basic methods to solve quadratic equations algebraically:

Given quadratic equation:

[tex]h(t)=-0.07t^2+0.007t+5[/tex]

Method 1 - Factoring

If the trinomial is able to be easily factored, set the equation equal to zero and factor.  Set each factor equal to zero and solve.

As the given example is not easily factored, factoring is not the best method to use on this occasion.

Method 2 - Completing the Square

If the quadratic expression is not able to be easily factored, try completing the square.

Divide all terms by [tex]a[/tex] (-0.07):

[tex]\implies t^2-\dfrac{1}{10}t-\dfrac{500}{7}=0[/tex]

Move the constant to the right side:

[tex]\implies t^2-\dfrac{1}{10}t=\dfrac{500}{7}[/tex]

Add the square of half the coefficient of t to both sides:

[tex]\implies t^2-\dfrac{1}{10}t+\left(-\dfrac{1}{20}\right)^2=\dfrac{500}{7}+\left(-\dfrac{1}{20}\right)^2[/tex]

[tex]\implies t^2-\dfrac{1}{10}t+\dfrac{1}{400}=\dfrac{200007}{2800}[/tex]

Factor the perfect trinomial:

[tex]\implies \left(t-\dfrac{1}{20}\right)^2=\dfrac{200007}{2800}[/tex]

Square root both sides:

[tex]\implies t-\dfrac{1}{20}=\pm\sqrt{\dfrac{200007}{2800}[/tex]

Add 1/20 to both sides:

[tex]\implies t=\dfrac{1}{20}\pm\sqrt{\dfrac{200007}{2800}[/tex]

As decimals to 2 decimal places:

[tex]\implies t=8.50, -8.40\:\:\sf(2 \:d.p.)[/tex]

Method 3 - Quadratic Formula

[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]

This can be used to find both real and imaginary or complex solutions.

Identity the variables:

[tex]a=-0.07, \quad b=0.007, \quad c=5[/tex]

Substitute the values into the quadratic formula and solve for t:

[tex]\implies t=\dfrac{-0.007 \pm \sqrt{0.007^2-4(-0.07)(5)} }{2(-0.07)}[/tex]

[tex]\implies t=\dfrac{-0.007 \pm \sqrt{1.400049}}{-0.14}[/tex]

[tex]\implies t=\dfrac{0.007 \pm \sqrt{1.400049}}{0.14}[/tex]

[tex]\implies t=8.50, -8.40\:\: \sf (2 \:d.p.)[/tex]

Finally, a non-algebraic method of solving a quadratic equation is by graphing using a graphical calculator (see attached).  The solutions are the points at which the curve intercepts the x-axis.

For Janna to prove that △ABC∼△DEF by the SSS similarity theorem, she will use: B. DE/AB = EF/BC = DF/AC = 1/4.

According to the SSS similarity theorem, if the ratio of the corresponding sides of two triangles are equal, then both triangles are similar.

The ratio of the corresponding sides of the triangles given are:

DE/AB = 4/16 = 1/4

EF/BC = 3.25/13 = 1/4

DF/AC = 1.75/7 = 1/4

Therefore, what Janna needs is:

B. DE/AB = EF/BC = DF/AC = 1/4

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Answer: 9

Step-by-step explanation:

The interior angle of the 8-sided polygon is

[tex]\frac{180(8-2)}{8}=135^{\circ}[/tex]

The interior angle of the 10-sided polygon is

[tex]\frac{180(10-2)}{10}=144^{\circ}[/tex]

Taking the difference, we get [tex]b=9^{\circ}[/tex]

The pair of equations that generates graphs with the same vertex is

B) y = –4x2 and y = 4x2

A vertex serves as  angular point that allows the meeting of two or more lines .

To express a quadratic equation we can use :

[tex][y = a(x - h) ^{2} + k][/tex]

where vertex is (-h , k)

If we look at first option, the the vertices are  [tex](-4 , 0) and (4 , 0)[/tex]

Then the second option, graphs' vertices are the same thing which is ,[tex](0 , 0).[/tex]

Therefore, the pair of equations that generates graphs with the same vertex is B) y = –4x2 and y = 4x2

CHECK THE COMPLETE QUESTION BELOW:

Which pair of equations generates graphs with the same vertex?

A) y = –(x + 4)2 and y = (x – 4)2

B) y = –4x2 and y = 4x2

C) y = –x2 – 4 and y = x2 + 4

D) y = (x – 4)2 and y = x2 + 4

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Answer:

294.84 ft²

Step-by-step explanation:

A = (B + b)h/2

A = (28.5 ft + 18.3 ft)(12.6 ft)/2

A = 294.84 ft²

Answer: 294.84 ft^2

Step-by-step explanation:

Area of trapezoid=((a+b)/2)*h

here

a=28.5 ft

b=18.3 ft

h=12.7 ft

Area=((28.5+18.3)/2)*(12.7)

Area=294.84 ft^2

The location of the fountain based on the information given is (-1, 0.5).

From the graph, the coordinates are (4, 2) and (-4,   -2).

The ratio is given as 5:3.

The coordinates of the fountain will be:

[(5 × -4 + 3 × 4)/(5 + 3), (5 × -3 + 3 × 2)/(5 + 3)]

= (-1, -0.5)

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