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3 years ago

Proving lines parallel!!! please help

Mathematics

1 answer:

Fynjy0 [20]3 years ago

4 0

Answer:

Step-by-step explanation:

Since C and A are parallel

and B is perpendicular to C

then B is also perpendicular to A

meaning they have the same angle so 90 Degrees.

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Answer both please asap (there’s a picture)

krek1111 [17]

Answer:

(0, 1)

Step-by-step explanation:

So, the spot on the graph where they cross.

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3 years ago

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I NEED HELP PLEASE !!!! ;(

marissa [1.9K]

Answer:

The answer is always

Hope this helps......... :)

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3 years ago

Which expression is equivalent to (81x4y16)12 for all positive values of x and y?<br> 40.5x4y16

IrinaVladis [17]

Answer:

$(81x^4y^{16})^{\frac{1}{2}} = 9 x^2 y^8$

Step-by-step explanation:

Given

$(81x^4y^{16})^{\frac{1}{2}}$

Required

Determine an equivalent expression

Express 81 as $9^2$

$(9^2x^4y^{16})^{\frac{1}{2}}$

Apply law of indices

$9^2^{\frac{1}{2}} * x^4^{\frac{1}{2}} * y^{16}^{\frac{1}{2}}$

$9^{\frac{2*1}{2}} * x^{\frac{4*1}{2}} * y^{\frac{16*1}{2}}$

$9^{\frac{2}{2}} * x^{\frac{4}{2}} * y^{\frac{16}{2}}$

$9^1 * x^2 * y^8$

$9 * x^2 * y^8$

$9 x^2 y^8$

Hence:

$(81x^4y^{16})^{\frac{1}{2}} = 9 x^2 y^8$

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2 years ago

Priya bought these items at the grocery store. Find each unit price. A. 12 eggs for $3. How much is the cost per egg? B. 3 pound

Ratling [72]

Answer: A. Cost per egg = $4 .

B. Cost per pound = $ 2.5

C. Cost per roll = $ 0.5

D. Cost per apple = $0.35

Step-by-step explanation:

Formula: cost per item = (total cost) ÷ (number of items)

A. 12 eggs for $3.

Cost per egg = $ (12÷3) = $4

So, Cost per egg = $4 .

B. 3 pounds of peanuts for $7.50

Cost per pound of peanuts = $ (7.50 ÷ 3)= $ 2.5

So, Cost per pound = $ 2.5

C. 4 rolls of toilet paper for $2.

Cost per roll = $ (2 ÷4 ) = $ 0.5

So, Cost per roll = $ 0.5

D. 10 apples for $3.50.

Cost per apple = $ (3.50 ÷ 10)= $ 0.35

So, Cost per apple = $0.35

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3 years ago

Plz help me I'm too Du-mb

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

You must have x alone

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3 years ago

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