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2 years ago

Q bisects PR. If PR=28, and PQ = 2x+5, what is the value of x?

Mathematics

1 answer:

Veseljchak [2.6K]2 years ago

3 0

Answer:

$x = 4.5$

Step-by-step explanation:

Given

$PR = 28$

$PQ = 2x + 5$

Required

Find x

Since Q is a bisector of PR; This mean PR is divided into exactly two equal parts;

Hence:

$PR = 2 * PQ$

Substitute values for PR and PQ

$28 = 2 * (2x + 5)$

Divide both sides by 2

$\frac{1}{2} * 28 =\frac{1}{2} * 2 * (2x + 5)$

$14 = 2x + 5$

Solve for 2x

$2x = 14 - 5$

$2x = 9$

Solve for x

$x = \frac{9}{2}$

$x = 4.5$

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10 POINTS<br> Find x. Assume that segments that appear tangent are tangent.

mixas84 [53]

Answer:

Step-by-step explanation:

see explanation, use tangent property to solve question

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3 years ago

Mike is working on solving the exponential equation 37^x = 12; however, he is not quite sure where to start.

Rainbow [258]

The solution of the given exponential equation is 0.688.

Given that Mike is working on solving the exponential equation 37ˣ = 12.

An exponential equation is an exponential equation where the power (or) part of the exponent is a variable.

firstly, we have to slve this equation is by converting it to logarithmic form. Any exponential equation can be transformed into an equivalent logarithmic equation as follows:

aˣ = y

logₐy = x

Now, we will apply this transformation to our equation and we get

log₃₇12=x

Further, we will apply the change of base formula so that solution is written in terms of base 10 logs:

x=log12/log37

So, this is an exact answer to given equation, but we can simplify it further by using decimal approximation of it using a calculator. Remember that these logs are base 10:

x≈1.079/1.568

x=0.688

Hence, the solution of the given exponential equation 37ˣ=12 is 0.688.

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1 year ago

Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are inches long, inches wide, and 8 inche

allochka39001 [22]

Answer:

- The number of small boxes that will fill the large box 1 = 64

- The number of small boxes that will fill the large box 2 = 56

Step-by-step explanation:

Complete Question

Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are 4.5 inches long, 4.5 inches wide, and 8 inches tall. To transport the items to an event, you want to know how many of the small boxes will fit in larger boxes. The larger boxes are available in two sizes. Large Box 1 is 24.25 inches long, 18 inches wide, and 24 inches tall. Large Box 2 is 20.5 inches long, 18.5 inches wide, and 24 inches tall. Both the small and large boxes must remain upright.

Solution

To know how many of the small boxes will fit in larger boxes, we need to obtain the volumes of the small box, large box 1 and large box 2.

Volume of a cuboid = L × W × H

For the small box,

Length = L = 4.5 inches

Width = W = 4.5 inches

Height = H = 8 inches

Volume of the small box = 4.5 × 4.5 × 8 = 162 in³

For large box 1,

Length = L = 24.25 inches

Width = W = 18 inches

Height = H = 24 inches

Volume of the large box 1 = 24.25 × 18 × 24 = 10,476 in³

For large box 2

Length = L = 20.5 inches

Width = W = 18.5 inches

Height = H = 24 inches

Volume of the large box 2 = 20.5 × 18.5 × 24 = 9,102 in³

The number of small boxes that'll fill the large box 1 = (10,476/162) = 64.667 = 64 small boxes (rounded down because the fraction cannot be forced into the large box 1.

The number of small boxes that will fill the large box 2 = (9,102/162) = 56.185 = 56 small boxes.

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2 years ago

Consider the following system of equations:

Delicious77 [7]

Answer:

The system has infinitely solutions

Step-by-step explanation:

we have

$-\frac{1}{3}x^{2}=-\frac{5}{6}+\frac{1}{3}y^{2}$

$\frac{1}{3}x^{2}+\frac{1}{3}y^{2}=\frac{5}{6}$

Multiply by 3 both sides

$x^{2}+y^{2}=\frac{5}{2}$ ----> equation A

The equation A is a circle centered at origin with radius $r=\sqrt{5/2}\ units$

and

$5y^{2} =\frac{25}{2}-5x^{2}$

$5x^{2}+5y^{2} =\frac{25}{2}$

Divide by 5 both sides

$x^{2}+y^{2} =\frac{5}{2}$ ----> equation B

The equation B is a circle centered at origin with radius $r=\sqrt{5/2}\ units$

Equation A and Equation B are the same

Therefore

The system has infinitely solutions

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2 years ago

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Mamont248 [21]

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Step-by-step explanation:

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