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3 years ago

Solve for x: 5(x-2)=2(10 x)

Mathematics

1 answer:

OverLord2011 [107]3 years ago

6 0

Distribute
5x-10=20+2x
minus 2x both sides
3x-10=20
add 10 both sides
3x=30
divide both sides by 3
x=10

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Oli's stuffed bear shop has 456 bears in stock. Exactly half of them are made from wool. What is the divisor

tigry1 [53]

2 is the divisor and 456 is the dividend. You want to find a way to get half, therefore you will divide the total bears (456) by 2.
456÷2= 228

4 0

3 years ago

Which equation below gives an incorrect value for the function k(x) = 512x?

lina2011 [118]

Answer:

D. gives an incorrect value for the function k(x) = $512^{x}$

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.

Which equation below gives an incorrect value for the function k(x) = $512^{x}$ ?

A.) k(2/3)=64

B.) k(4/9)=16

C.) k(1/3)=8

D.) k(2/9)=8

My answer:

Let analyse all 4 possible answers:

A.) k(2/3)=64

<=> k(2/3) = $512^{2/3}$ = 64

True

B.) k(4/9)=16

<=> k(4/9) = $512^{4/9}$ = 16

True

C. k(1/3)=8

<=> k(1/3)= $512^{1/3}$ = 8

True

D. k(2/9)=8

<=> k(2/9) = $512^{2/9}$ = 4 not 8

Wrong

Hope it will find you well

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3 years ago

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monitta

Answer:

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6 0

3 years ago

Read 2 more answers

A swimming pool is being filled at the rate of 12 cubic yards per minute. If the pool is 18 yards long, 10 yards wide, and 3 yar

horsena [70]

(18*10*3yd^3)(min/12yd^3)

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7 0

4 years ago

Read 2 more answers

For the concentric circles, the outer circle radius is r=6 in while that of the inner circle is r=3 in. Find the exact area of t

worty [1.4K]

Answer:

The exact area of the annulus (ring) made by the two circle is $27\pi \ in^{2}$

Step-by-step explanation:

Given:

Let the radius of outer circle i.e CA be $r_{o}= 6\ in$

Let the radius of inner circle i.e CB be $r_{i}= 3\ in$

The diagram is given below as attachment.

$\textrm{area of circle}= \pi r^{2} \\\textrm{area of the shaded region} =\textrm{area of outer circle}-\textrm{area of inner circle}\\\textrm{area of the annulus ring}=\pi r_{o}^{}2 - \pi r_{i}^{}2$

Substituting the values we get

$\textrm{area of the annulus ring}=\pi\times 6^{2} - \pi\times 3^{2}\\=\pi (36-9)\\=27\pi\ in^{2}$

4 0

4 years ago