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r-ruslan [8.4K]
3 years ago
13

What is the domain of this graph? ​

Mathematics
1 answer:
sweet [91]3 years ago
6 0

Answer:

the domain is all real numbers

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Need help with this one as well. I wasn't sure so I went with C.
disa [49]

convergent is |r|>1

so the answer would have to be less that 1.

answer is A.


6 0
3 years ago
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
260% of blank is 156
Reptile [31]

Answer:60?

Step-by-step explanation:

260%=2.6

156/2.6

5 0
4 years ago
Analyze the diagram below and complete the instructions that follow.<br> Find the value of x.
LUCKY_DIMON [66]

Answer:

10

Step-by-step explanation:

since it is a right angle triangle

{x}^{2}  =  ({5 \sqrt{2} })^{2}  +({5 \sqrt{2} })^{2}

x =  \sqrt[2]{(5 \sqrt{2} )^{2}  +( 5 \sqrt{2})^{2}}

x =  \sqrt{(25 \times 2) + (25 \times 2)}

x =  \sqrt{50  + 50}

x =  \sqrt{100}

x = 10

3 0
3 years ago
Find the product of -6/9 and it's multiplication inverse
Sladkaya [172]

Answer:

did that help <<<<<<<<<<

7 0
3 years ago
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