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2 years ago

Find the Value of each power.

Mathematics

1 answer:

Alexxandr [17]2 years ago

4 0

Answer:

144

8/343

0.16

4/9

0.343

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There were 60 seventh grade students who signed up for soccer tryouts last year. This year, 48 seventh graders signed up for try

Lubov Fominskaja [6]

Answer:

20%

Step-by-step explanation:

(60 - 48) / 60 = 0.2 (20%)

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3 years ago

You have an ice cream cone that is 6 inches tall with a radius of 2 inches. The cone is completely filled with ice cream. There

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3 years ago

Need help with question 15! please!!! This is due 02/2/22 which is tmr please!

Airida [17]

Answer: x = 6

Step-by-step explanation:

Because we are given that the two figures are already similar, we can set up a proportional relationship to determine the value of x. Since the larger rectangle has a width of 21 and the smaller rectangle has a width of 7, those values can be used as the numerators in the proportion. Since the larger rectangle's length is x plus the additional 3 units on the smaller rectangle, x+3 and 3 can be used as the denominators in the proportion.

Next, we set up the proportion as follows:

$\frac{21}{x+3} =\frac{7}{3}$

Then, you cross multiply to get the equation:

$63=7x+21$

After solving down algebraically, you get the value of x to be 6.

I hope this helps!

5 0

2 years ago

Item 5

zvonat [6]

Answer:

The answe would be "All parallelogram are quadrilaterals."

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2 years ago

If alpha and beta are the angles in the first quadrant tan alpha = 1/7 and sin beta =1/ root 10 then usind the formula sin (A +B

zysi [14]

Answer:

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

Step-by-step explanation:

$\alpha \text{ and } \beta \text{ in Quadrant I}$

$\tan(\alpha)=\frac{1}{7} \text{ and } \sin(\beta)=\frac{1}{\sqrt{10}}=\frac{\sqrt{10} }{10}$

<u>Using Pythagorean Identities</u>:

$\boxed{\sin^2(\theta)+\cos^2(\theta)=1} \text{ and } \boxed{1+\tan^2(\theta)=\sec^2(\theta)}$

$\left(\frac{\sqrt{10} }{10} \right)^2+\cos^2(\beta)=1 \Longrightarrow \cos(\beta)=\sqrt{1-\frac{10}{100}} =\sqrt{\frac{90}{100}}=\frac{3\sqrt{10}}{10}$

$\text{Note: } \cos(\beta) \text{ is positive because the angle is in the first qudrant}$

$1+\left(\frac{1 }{7} \right)^2=\frac{1}{\cos^2(\alpha)} \Longrightarrow 1+\frac{1}{49}=\frac{1}{\cos^2(\alpha)} \Longrightarrow \frac{50}{49} =\frac{1}{\cos^2(\alpha)}$

$\Longrightarrow \frac{49}{50}=\cos^2(\alpha) \Longrightarrow \cos(\alpha)=\sqrt{\frac{49}{50} } =\frac{7\sqrt{2}}{10}$

$\text{Now let's find }\sin(\alpha)$

$\sin^2(\alpha)+\left(\frac{7\sqrt{2} }{10}\right)^2=1 \Longrightarrow \sin^2(\alpha) +\frac{49}{50}=1 \Longrightarrow \sin(\alpha)=\sqrt{1-\frac{49}{50}} = \frac{\sqrt{2}}{10}$