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3 years ago

PLSSS HURRY!!!

Mathematics

2 answers:

aniked [119]3 years ago

8 0

Answer:

<h2>B. Clockwise Rotation</h2>

Step-by-step explanation:

i hope it helps ;)

Vlad1618 [11]3 years ago

6 0

B. Cya adrianna ps. U shouldn’t include ur name in the photos

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Identify the constant term in this expression 7p - 3p + 9 + x

ANEK [815]

Answer:

the constant in this expression is 9 because it is not a coefficient of any variable

7 0

3 years ago

Read 2 more answers

When 1,250 Superscript three-fourths is written in simplest radical form, which value remains under the radical?

Dvinal [7]

The value remains under the radical is 8 ⇒ last answer

Step-by-step explanation:

Let us revise how to write the exponent as a radical

$a^{\frac{m}{n}}$ can be written as $\sqrt[n]{a^{m}}$
To simplify the radical factorize the base "a" to its prime factors

Example:

$(54)^{\frac{2}{3}}=\sqrt[3]{(54)^{2}}$ ,
Factorize 54 into prime factors ⇒ 54 = 2 × 3 × 3 × 3 = $2(3)^{3}$
$\sqrt[3]{(54)^{2}}=\sqrt[3]{[2(3^{3}]^{2}}=\sqrt[3]{2^{2}*3^{6}}$
2² can not go out the radical because 2 is less than 3 not divisible by 3
$3^{6}$ can go out the radical because 6 is divisible by 3, then divide 6 by 3, so it will be 3² out the radical
$\sqrt[3]{(54)^{2}}=3^{2}\sqrt[3]{2^{2}}=9\sqrt[3]{4}$

Now let us solve your problem

∵ $1250^{\frac{3}{4}}=\sqrt[4]{1250^{3}}$

- Factorize 1250 to its prime factors

∵ 1250 = 2 × 5 × 5 × 5 × 5

∴ $1250=2*5^{4}$

∴ $\sqrt[4]{(2*5^{4})^{3}}=\sqrt[4]{2^{3}*5^{12}}$

∵ 2³ can not go out the radical because 3 < 4 and not divisible by it

- $5^{12}$ can go out the radical because 12 can divided by 4

∵ 12 ÷ 4 = 3

∴ $5^{12}$ can go out the radical as 5³

∴ $\sqrt[4]{1250}=5^{3}\sqrt[4]{2^{3}}$

∴ $\sqrt[4]{1250}=125\sqrt[4]{8}$

∴ The value remains under the radical = 8

The value remains under the radical is 8

Learn more:

You can learn more about the radicals in brainly.com/question/7153188

#LearnwithBrainly

8 0

3 years ago

X+6y=28, solve for y

m_a_m_a [10]

Answer:

$y = -\frac{x}{6} + \frac{14}{3}$

Step-by-step explanation:

So when you're solving for y you want to get y by it self meaning there should be nothing with y, including any coefficients. The first step is to usually move any terms to the other side and then divide by the coefficient of y so y "technically doesn't" have a coefficient, while it does, but that coefficient 1, so it doesn't really mean much.

Subtract x from both sides

$6y = -x+28$