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Umnica [9.8K]
3 years ago
13

A small hole in the wing of a space shuttle requires a 20.7 cmଶ patch. What is the patch’s area in square kilometers (kmଶ )? If

the patching material costs NASA $3.25 inଶ ⁄ , what is the cost of the patch?
Mathematics
1 answer:
Reptile [31]3 years ago
3 0

Answer:

look it up

Step-by-step explanation:

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Cot0 + tan0 = sec0csc0
Lelechka [254]

Answer:

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Step-by-step explanation:

7 0
3 years ago
4/5 divided by 5 1/2
irina1246 [14]

Answer:

8/55

Step-by-step explanation:

4/5 DIVIDED 5 1/2 = 8/55

6 0
3 years ago
Read 2 more answers
HELPP!!!!!!!!!!!!!!!!
Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step.  The equation is

8-3\sqrt[5]{x^3}=-7

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time.  First thing is to subtract 8 from both sides:

-3\sqrt[5]{x^3}=-15

The goal is to isolate the term with the x in it, so that means that the -3 has to go.  Divide it away on both sides:

\sqrt[5]{x^3}=5

Let's rewrite that radical into exponential form:

x^{\frac{3}{5}}=5

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely.  On the right, let's rewrite that back in radical form to solve it easier:

x=\sqrt[3]{5^5}

Let's group that radicad into groups of 3's now to make the simplifying easier:

x=\sqrt[3]{5^3*5^2} because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

x=5\sqrt[3]{5^2} which is the same as:

x=5\sqrt[3]{25}

8 0
3 years ago
Decide whether it’s a Function or not
cupoosta [38]
1=function
2=not function
3=not function
4=function
8 0
1 year ago
Amy jogs 1/3 of a mile in 1/15 of an hour, while john takes 1/30 of an hour to jog 1/5 of a mile. If they continue at this rate,
jolli1 [7]

Step-by-step explanation:

Given that,

Amy jogs 1/3 of a mile in 1/15 of an hour and john takes 1/30 of an hour to jog 1/5 of a mile.

Speed of Amy,

v_1=\dfrac{\dfrac{1}{3}\ miles }{\dfrac{1}{15}\ h}=5\ mph

Speed of John,

v_2=\dfrac{\dfrac{1}{5}\ miles }{\dfrac{1}{30}\ h}=6\ mph

John will jog farther in one hour as he is moving with faster speed. The speed of John is 1 mph more than Amy.

6 0
3 years ago
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