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Y_Kistochka [10]
2 years ago
7

Can you please solve this​

Mathematics
1 answer:
ddd [48]2 years ago
7 0
The answer to your question is a=124 because the center triangle is a isosceles triangle
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Is 4/6 = 9/12 not proportionate
andre [41]

\text{If}\ \dfrac{a}{b}=\dfrac{c}{d}\ \text{is proportional, then}\ ad=bc.\\\\\text{We have}\ \dfrac{4}{6}=\dfrac{9}{12}.\ \text{Cross multiply}\\\\(4)(12)=(6)(9)\\\\48=54\ FALSE\\\\Answer:\ NOT\ PROPORTIONAL

4 0
2 years ago
Simplify the following expression 4^-11/3÷4^-2/3<br>A.1/12<br>B.1/64<br>C.12<br>D.64​
Goryan [66]

Question:

Simplify the following expression:

4^{-\frac{11}{3}÷4^{-\frac{2}{3} }

A.1/12

B. 1/64

C.12

D.64

Answer:

B. 1/64

7 0
3 years ago
Ned created a video that was originally 40 minutes
Zanzabum

Answer:

36.8 minutes or 36 minutes 48 seconds

Step-by-step explanation:

8 percent of 40 is 3.2

(.08 times 40)

40 minus 3.2 is 36.8 :)

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3 0
3 years ago
Read 2 more answers
if a cylinder with height 9 inches and radius r is filled with water, it can fill a certain pitcher. how many of these pitchers
dalvyx [7]

Answer:

4 pitchers

Step-by-step explanation:

we know that

The volume of a cylinder is equal to

V=\pi r^{2}h

step 1

Find the volume of the cylinder  with height 9 inches and radius r

substitute the given values

V_1=\pi r^{2}(9)

V_1=9\pi r^{2}\ in^3

step 2

Find the volume of the cylinder  with height 9 inches and radius 2r

substitute the given values

V_2=\pi (2r)^{2}(9)

V_2=36\pi r^{2}\ in^3

step 3

we know that

The volume of the cylinder 1 can fill a certain pitcher,

so

The volume of the pitcher is the same that the volume of cylinder 1

therefore

the number of pitchers that can be filled by the second cylinder is equal to divide the volume of the second cylinder by the volume of the first cylinder

\frac{36\pi r^{2}}{9\pi r^{2}}=4\ pitchers

3 0
3 years ago
If x = 6, which equation is true?
Mandarinka [93]

Answer:

Sorry, I can't help you because you didn't provide enough information.

Step-by-step explanation:

4 0
3 years ago
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