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Naily [24]
3 years ago
14

How can u answer this question in words by the shadow u can tell it is 9 dark outside PLZ help

Mathematics
1 answer:
valentinak56 [21]3 years ago
5 0
Ok. So remember those questions you got in kindergarten example: 40 people were asked which was the nicest fruit in their opinion and here is a graph of what they thought was the nicest. So people use number bars as a way to organize problems. Does this make any sense? I will help in comments if it doesn't.

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Alenkasestr [34]

Answer:

Step-by-step explanation:

b. The set of all pairs of line segments c. Congruent pairs of segments have the same measure. 8. ... ZB ZDCA,. 2. Given. ZA. ZDCB. 3. mZB mZDCB,. 3. Definition of. mZA. mZDCB congruent angles .

7 0
3 years ago
Help please! I don't understand how to do this at all. Please help
Viktor [21]
9. \sqrt[n]{x}
    \sqrt[6]{-64}
    \sqrt[6]{64 * (-1)}
    \sqrt[6]{64}\sqrt[6]{-1}
    2i

10. \sqrt[n]{x}
      \sqrt[3]{128}
      \sqrt[3]{64 * 2}
      \sqrt[3]{64}\sqrt[3]{2}
      4\sqrt[3]{2}

11. \sqrt[n]{x}
      \sqrt[10}{1024}
      2

12. \sqrt[n]{x}
      \sqrt[7]{\frac{8\sqrt{2}}{2187}}
      \frac{\sqrt[7]{8\sqrt{2}}}{\sqrt[7]{2187}}
      \frac{\sqrt[7]{8\sqrt{2}}}{\sqrt[7]{(3)^{7}}}
      \frac{\sqrt[7]{8\sqrt{2}}}{3}
      
4 0
3 years ago
M: Mental
katrin [286]

Answer:

Water is not wet, it makes things wet.

5 0
3 years ago
Read 2 more answers
If a right triangle's hypotenuse is 17 units long, and one of its legs is 15 units long, how long is the other leg?
Vesna [10]

Answer:

8 units

Step-by-step explanation:

Hello!

So, there's a formula we can apply to right-angled triangles: Pythagorean's theorem. It states that  c = \sqrt{{a}^2 + b^{2} }, where <em>c</em> is the hypotenuse and <em>a</em> and <em>b </em>are the legs of the triangle.

So, from the problem,  if <em>c </em>= 17 and <em>a </em> = 15, then, we're solving for <em>b</em>. So we'll rewrite the theorem to solve for <em>b</em>.

{c}^2 = {a}^2+{b}^2\\{c}^2-{a}^2={b}^2\\{b} = \sqrt{{c}^2-{a}^2}

Okay, so now we have isolated the theorem for <em>b. Let's </em>plug in our values for <em>c </em>and <em>a</em>.

b = \sqrt{{17}^2-{15}^2}\\b = \sqrt{289-225}\\b = \sqrt{64}\\b = 8

So, using the theorem, we found <em>b</em> = 8. To check our work, let's plug in <em>b</em> and <em>a</em> and solve for <em>c.</em>

<em />c = \sqrt{{a}^2+{b}^2}}\\c = \sqrt{{15}^2+{8}^2}\\c = \sqrt{225+64}\\c = \sqrt{289}\\c = 17\\<em />

So, we got our hypotenuse to equal 17 units, which is correct! So, our <em>b</em> is correct too. Awesome

7 0
3 years ago
HELPPPPPPP
quester [9]

Answer:

It's A. (2, 0) and (0, -4)

3 0
2 years ago
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