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3 years ago

6

A +b have played table tennis against each other 30 times. The ratio of the number of times A has won to the number of times B h

as won is 3:7. How many times has B won?

1 answer:

viva [34]3 years ago

3 0

B won bc it has played the mind

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Someone please please help me it’s Algebra

tankabanditka [31]

Answer:

A shift to the right 8 units

Step-by-step explanation:

Remember to take the opposite of the sign when you are moving horizontally. The -8 signifies that it will be moving right 8 units instead of left.

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3 years ago

Siri can run 6 miles in 3/4 of an hour at this rate what is the total number of miles a student can run in one hour

Lesechka [4]

Answer:

4.5 mph

Step-by-step explanation:

6 times 3 is 18.

18 divided by 4 is 4.5

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3 years ago

12. Write the first five terms of two different sequences that have 10 as the

kodGreya [7K]

Answer:

a) The arithmetic sequence with common difference 2 that has 8 as the first term.

b) The arithmetic sequence of common difference -5 and first term 15.

Step-by-step explanation:

Let's use for example the arithmetic sequence with common difference 2 that has 8 as the first term. Then the first two terms of this sequence are:

8, and (8+2) = 10 Therefore the second term is 10.

Another arithmetic sequence of common difference -5 and first term 15. The firs two terms of this sequence are:

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6 0

3 years ago

What is the following sum 3b^2

densk [106]

The given expression is $3b^2*(\sqrt[3]{54a}) + 3*(\sqrt[3]{2ab^6})$

This can be simplified as :

= $3*b^2*(\sqrt[3]{27 *2*a}) + 3*(\sqrt[3]{2*a*b^6})$

We know that: $\sqrt[3]{27} = 3$

Similarly we also can simplify: $\sqrt[3]{b^6} = b^2$

So our expression will look like this:

= $3*3*b^2*(\sqrt[3]{2a}) + 3*b^2*(\sqrt[3]{2a})$

= $9b^2*(\sqrt[3]{2a}) + 3b^2*(\sqrt[3]{2a})$

= $\sqrt[3]{2a}*(9b^2 + 3b^2)$

= $\sqrt[3]{2a}*(12b^2)$

This can also be written as:

$12b^2(\sqrt[3]{2a})$

So the Answer is Option B

6 0

3 years ago

Read 2 more answers

FIRST CORRECT ANSWER GETS BRAINLIEST AND 50 POINTS.

AnnyKZ [126]

Answer:

A) $5^3*5^{-4}=5^{3-4}=5^{-1}=\frac{1}{5}$ (Yes. It has a value between zero and one.)

B) $\frac{3^5}{3^{-6}}=3^{5-\left(-6\right)}=3^{5+6}=3^{11}=177147$ (No. It does not have a value between zero and one.)

C) $\left(\frac{1}{4}\right)^3* \left(\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^{3+2}=\left(\frac{1}{4}\right)^5=\frac{1^5}{4^5}=\frac{1}{1024}$ . (Yes. It has a value between zero and one.)

D) $\frac{\left(-7\right)^5}{\left(-7\right)^7}=\left(-7\right)^{5-7}=\left(-7\right)^{-2}=\frac{1}{\left(-7\right)^2}=\frac{1}{49}$ (Yes. It has a value between zero and one.)

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3 years ago

Read 2 more answers