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2 years ago

PLEASE ANSWER ASAP!!! I HAVE 2 HRS!!!

Mathematics

1 answer:

garik1379 [7]2 years ago

6 0

Answer:

Step-by-step explanation:

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Simplify (√2)(3^√2)

In-s [12.5K]

Answer:

√2·3^√2

Step-by-step explanation:

5 0

3 years ago

when you're dividing fractions you only flip the numbers around and multiply them if the numerator is higher than the denominato

Whitepunk [10]

Answer:

Step-by-step explanation:

Actually, when dividing fractions, you always flip the second fraction and multiply, no matter the size of the numerator or denominator.

6 0

2 years ago

The sum of two numbers is -10.5.

grandymaker [24]

Answer:

-5 and -5.5;

-12.5 and 2

Step-by-step explanation:

Two negative addends, can be added together to give -10.5.

For example:

(-5) + (-5.5) = -5 - 5.5 = -10.5

Also, it is possible for one of the addends to be negative while the other is positive, and their sum will give us -10.5.

For example:

The sum of -12.5 and 2 will give us -10.5.

We are adding a positive and a negative number here. As usual, we will subtract the smaller number from the bigger number, while the result will carry the sign of the bigger number, which in this case is negative sign.

Thus:

(-12.5) + (2) = -10.5

4 0

3 years ago

A rectangle has an area of 616 m² and a width of 22m. What is its length?

daser333 [38]

Answer:

l=28m

w Width

A Area

616

m²

Using the formula

A=wl

Solving forl

l=A

w=616

22=28m

hope this will help you

make me brainliest

8 0

2 years ago

Read 2 more answers

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago