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katovenus [111]
3 years ago
5

Plz help me with this problem!

Mathematics
1 answer:
sergiy2304 [10]3 years ago
7 0

Answer:

It's 99.00900

Step-by-step explanation:

Hope it helps you.......

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Please please help needs to be submitted in 15 minutes!!!
ollegr [7]

Answer:

Volume of prism 1: 104 cubic inches

Surface area of prism 2: 650 square inches

Step-by-step explanation:

To find the volume of prism 1, set up a proportion using the side lengths and the volume of prism 2.

The ratio of the side lengths will be your scale factor. When finding volume using scale factor, you need to cube the scale factor. The numerator in the proportion are from prism 1 and the denominator is prism 2. Prism 1 is the numerator since it is the new image and we are solving for it's volume.

(\frac {4}{10}) ^{3}  =\frac{x}{1625} \\\\\frac{64}{1000} =\frac{x}{1625} \\\\

1000x=104000\\\frac{1000x}{1000} =\frac{104000}{1000} \\x = 104

To find the surface area of Prism 2, we are going to setup another proportion, but this time we are going to square the scale factor since we are finding area. Prism 2 will be the numerator and prism 1 will be the denominator.

(\frac{10}{4} )^2 = \frac{x}{104} \\\frac{100}{16} =\frac{x}{104} \\16x=10400\\\frac{16x}{16}=\frac{10400}{16}\\ x= 650

6 0
2 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
Please help I will mark brainliest ITS DUE IN A FEW MINUTES PLS
AlladinOne [14]

Answer:

All real numbers except 0

Step-by-step explanation:

0 is the only number that's going to make it undefined (btw if you mark brainliest it would be appreciated :) ).

3 0
4 years ago
Which value is a solution of the equation 85= 120-w?
Ronch [10]
85=120-w\\85-120=120-w-120\\-35=-w\\\frac{-35}{-1}=\frac{-w}{-1}\\35=w

w = 35
8 0
3 years ago
Read 2 more answers
Interval points less than -1
grin007 [14]
Ini bahasa apa ya saya tidak tahu
4 0
3 years ago
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