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3 years ago

Which statements must be true?

Mathematics

2 answers:

xenn [34]3 years ago

7 0

Answer:

CB || XY

AB || XZ

CA= $\frac{1}{2}$ YZ (This wasn't an option you listed, but my test offered it)

Step-by-step explanation:

Verified correct with test results.

iVinArrow [24]3 years ago

5 0

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Please find the complete question in the attached file:

Using the formula for Mid-point:

$\bar{AB} || \bar{XZ} \\\\CA= \frac{1}{2} \ YZ\\\\\bar{CB} |||\bar{XY}$

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A car dealer's markup on every car they sell is 20%. For what price did the dealership buy a car that they sold for $18,600?

elena-s [515]

Answer:

buying price = $15,500

Step-by-step explanation:

selling price 20% more than the buying price

let the buying price be 100% then;

selling price = 120%

120% = $18,600

100% = ?

(100 × 18600) ÷ 120

= $15,500

8 0

3 years ago

Help, dividing fractions.

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3 years ago

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Alla [95]

0.002%. Just take 0.6/300 and input into a calculator.

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3 years ago

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gregori [183]

Answer:

use your distributive properties and combine the two equations and just use pemdas.

Step-by-step explanation:

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2 years ago

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a toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its heigh

sdas [7]

Answer:

t = 2.11 seconds

Step-by-step explanation:

A toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its height h at any given second t is modeled by the equation:

$h=-16t^2+32t+48$

Let us assume that we need to find the time by it to reach the ground. It means h = 0

$-16t^2+32t+48=0$

The above is a quadratic equation. The value of t is given by :

$t=\dfrac{-b\pm \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-2ac} }{2a},\dfrac{-b- \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-32+ \sqrt{(32)^2-2\times (-16)(8)} }{2\times (-16)},\dfrac{-32-\sqrt{(32)^{2}-2\times(-16)(8)}}{2\times(-16)}\\\\t=-0.11\ s, 2.11\ s$

So, it will take 2.11 seconds to reach the ground.

6 0

3 years ago