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3 years ago

4over 9 times 21 over six

Mathematics

1 answer:

Alex Ar [27]3 years ago

5 0

4/9 x 21/6 is this what you’re asking?
If so the answer is 1 5/9

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I need help with this V-4=v+2+7v-42 ??

zubka84 [21]

Answer:

$\boxed{\bold{v \ = \ \frac{36}{7}}}$

Step-by-step explanation:

Subtract v from both sides

$\bold{v-4-v=v+2+7v-42-v}$

Simplify

$\bold{-4=2+7v-42}$

Switch sides

$\bold{2+7v-42=-4}$

Group like terms

$\bold{7v+2-42=-4}$

Subtract the numbers: 2 - 42 = -40

$\bold{7v-40=-4}$

Add 40 to both sides

$\bold{7v-40+40=-4+40}$

Simplify

$\bold{7v=36}$

Divide both sides by 7

$\bold{\frac{7v}{7}=\frac{36}{7}}$

Simplify

➤ $\bold{v=\frac{36}{7}}$

5 0

3 years ago

I need to find the degree of this monomial: 4q^2 rs^6

Murljashka [212]

Answer:

The answer is 9.

Step-by-step explanation:

The degree of a monomial is defined as the sum of all the exponents of the variables, including the implicit exponents of 1 for the variables which appear without exponent.

2 + 6 + 1 = 9

Please mark me brainliest!

4 0

3 years ago

The triangles shown below must be congruent. right awnsers only please.

tankabanditka [31]

It is true. The triangles are congruent.

8 0

3 years ago

Read 2 more answers

How long will it take for an airplane that averages 95 miles an hour to fly 1,710 miles?

zysi [14]

The correct answer is 18 hope that helped♥

5 0

3 years ago

(ASAP PICTURE ADDED) What is the simplified form of the following expression?

bonufazy [111]

Answer:

option c is correct.

Step-by-step explanation:

$7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)$

WE need to simplify this equation.

Solve the parenthesis of each term.

$=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right$

Now, We will find factors of the terms inside the square root

factors of 2: 2

factors of 16 : 2x2x2x2

factors of 8: 2x2x2

Putting these values in our equation: $=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)$

Adding like terms we get:

$=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\$

$(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\can\,\,be \,\, written\,\, as\,\,\\(\sqrt[3] {2x})-6\left(\sqrt[3]{x}\right)$

So, option c is correct

5 0

3 years ago

Read 2 more answers