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Natali5045456 [20]
3 years ago
6

Can you help me with this? I don't seem to understand..

Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0
All you have to do is set up 2 proportions, where one paint you know and the other you do not know and make it equal to the ratio as 2 blue : 5 yellow

For example: 1 blue/x yellow = 2 blue/5 yellow, and then cross multiply and solve for x.

Do this for all unknowns, blue and yellow paint making it equal to the ratio and solve for the unknown.
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On a model castle, 1 inch equals 83 feet. The model is 10.5 inches wide. What is the actual width of the castle?
Furkat [3]

Answer:

871 and 1/2 ft.  

8 0
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If segment UV is considered to be the base of parallelogram TUVW, which segment would represent the altitude of the parallelogra
omeli [17]
The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>Segment TX is the segment that would best represent the altitude of the parallelogram.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
3 years ago
I need help on 12 please :)
alisha [4.7K]

Answer:

the answer is 10 to the 5th power

Step-by-step explanation:


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3 years ago
Turkey tastes like napkins
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4 0
3 years ago
A right triangle has side lengths of 0.6 meter and 0.8 meter. What is the length of the hypotenuse? Round to the nearest tenth.
Sonja [21]

Use the Pythagorean theorem since you are working with a right triangle:

a^2+b^2=c^2a2+b2=c2

The legs are a and b and the hypotenuse is c. The hypotenuse is always opposite the 90° angle. Insert the appropriate values:

0.8^2+0.6^2=c^20.82+0.62=c2

Solve for c. Simplify the exponents (x^2=x*xx2=x∗x ):

0.64+0.36=c^20.64+0.36=c2

Add:

1=c^21=c2

Isolate c. Find the square root of both sides:

\begin{gathered}\sqrt{1}=\sqrt{c^2}\\\\\sqrt{1}=c\end{gathered}1=c21=c

Simplify \sqrt{1}1 . Any root of 1 is 1:

c=c= ±11 *

c=1,-1c=1,−1

4 0
3 years ago
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