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3 years ago

GCF 70x^4,110x^5,60x^9

Mathematics

1 answer:

sweet [91]3 years ago

7 0

10x^4, i hope it helps

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Move words to the blanks to make the sentence true. Volume can be measured by counting the number of unit In a figure. cubes squ

nika2105 [10]

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•three-dimensional

Volume can be measured by number of unit CUBES in a THREE-DIMENSIONAL figure.

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3 years ago

Right triangle ABC is shown. Which equation can be used to solve for c?

julsineya [31]

Answer:

$\sin(50^\circ)=3/c$

Step-by-step explanation:

Remember SOH/CAH/TOA. We will use SOH, which means $\sin$ of the angle is equal to the opposite side over the hypotenuse.

6 0

3 years ago

Which set of three angles could represent the interior angles of a triangle? *

oksian1 [2.3K]

Use the triangle angle theorem:
All interior angles of a triangle equal 180 degrees.

If you use the theorem you should get the first one 19,70,91

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3 years ago

Someone help me on this please!

swat32

Answer:

x = 15 cm

Step-by-step explanation:

All the ratio of corresponding sides should be equal

10 corresponds to x

16 corresponds to 24

12 corresponds to 18

Finding the ratio of the known pairs:

$\frac{16}{24}=\frac{2}{3}$
$\frac{12}{18}=\frac{2}{3}$

Similarly 10 is to x should be 2 is to 3. Finding x:

$\frac{10}{x}=\frac{2}{3}\\10*3=2x\\30=2x\\x=\frac{30}{2}\\x=15$

So x = 15.

5 0

3 years ago

Read 2 more answers

Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is

ikadub [295]

The question is incomplete, the complete question is below:

Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is the temperature in Celsius at a height of kilometers above the planet's surface. The relationship is as follows: T(h) = 30.5 -2.5h.

a) Calculate T^-1(x)

b) T^-1(15)

Answer:

a) $T^{-1} (x)= \frac{30.5-x}{2.5}$

b) 6.2 °C

Step-by-step explanation:

a) $T(h) = 30.5 - 2.5h\\2.5h = 30.5-T(h)\\h = \frac{30.5-T(h)}{2.5}\\ T^{-1} (x)= \frac{30.5-x}{2.5}$

b) $T^{-1} (x)= \frac{30.5-x}{2.5}\\\\T^{-1} (15)= \frac{30.5-15}{2.5}=6.2$

3 0

3 years ago