Use <, >, or = to fill in the blanks. -7----- 0 Please help me!

Mathematics

1 answer:

Oduvanchick [21]2 years ago

3 0

Answer:

Step-by-step explanation:

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Pleaseeeeee helppppp, ill give you brainley i promise

Dima020 [189]

Its the third one for the answer

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2 years ago

How do you simplify on number 1?

Dennis_Churaev [7]

Okay. On the part that's not the whole number, see how the top number is BIGGER than the bottom? That means that it's not simplified. In order to make it simplified, you have to take a whole out of the pat that's not a whole number. Since the denominator is 12, the whole is 12/12. So, subtract it from the 13/12.

13/12-12/12=1/12

Now, since you found a whole inside of the fraction you have to add 1 to the whole number.
5+1=6

So, the answer is 6 1/12.

Thanks for the points!

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3 years ago

A resturant sells 10 tacos for 8.49,or 6 of the same kind of taco for 5.40.Which is a better deal

Aliun [14]

Answer:

10 tacos for 8.49 because 6 tacos for 5.40 it's almost a dollar for each.And the deal 10 tacos for 8.49 you pay more but it's better it's not a dollar for each it's like 84 cents for each

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3 years ago

Rewrite in simplest radical form 1/x^−3/6 Show each step of your process please!

Romashka [77]

$\frac{1}{a^{-b} } =a^b \\c^{ \frac{a}{b} }= \sqrt[b]{c^a} \\ \\ \frac{1}{x^{- \frac{3}{6} }}=\frac{1}{x^{- \frac{1}{2} }} =x^ \frac{1}{2} = \sqrt[2]{x^1} = \sqrt{x}$

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3 years ago

The functions f (theta) and g (theta) are sine functions, where f (0) equals g (0) equals 0.The amplitude of f (theta) is twice

777dan777 [17]

If period of $f(\theta)$ is one-half the period of $g(\theta)$ and
 $g(\theta)$ has a period of 2π, then $T_{g} =2T_{f}=2 \pi$ and $T_{f}= \pi$ .

To find the period of sine function $f(\theta)=asin(b\theta+c)$ we use the rule $T_{f}= \frac{2\pi}{b}$ .

f is sine function where f (0)=0, then c=0; with period $\pi$ , then $f(\theta)=asin 2\theta$ , because $T_{f}= \frac{2 \pi }{2} = \pi$ .

To find a we consider the condition $f( \frac{ \pi }{4} )=4$ , from where $asin2* \frac{\pi}{4} =a*sin \frac{ \pi }{2} =a=4$ .

If the amplitude of $f(\theta)$ is twice the amplitude of $g(\theta)$ , then $g(\theta)$ has a product factor twice smaller than $f(\theta)$ and while period of $g(\theta)$ is 2π and g(0)=0, we can write $g(\theta)=2sin\theta$ .

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2 years ago