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jekas [21]
2 years ago
11

Click image to see it all..... thank you ​

Mathematics
1 answer:
Semmy [17]2 years ago
7 0

Answer:

x = 7.5

Step-by-step explanation:

Proportions:

5            x

_     =     _

6            9

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Find the 58th term of the following arithmetic sequence.<br> 13, 19, 25, 31,
pashok25 [27]

Answer:

the answer to this is 355.

source:da brain

6 0
2 years ago
4x²-y²=171 and 2x-y=9, value of 2x+y
11111nata11111 [884]
So you will need to solve for x and y before evaluating 2x+y....

2x-y=9, y=2x-9  now this will make 4x^2-y^2=171 become:

4x^2-(2x-9)^2=171

4x^2-(4x^2-36x+81)=171

36x-81=171

36x=252

x=7, now we can use 2x-y=9 to solve for y...

2(7)-y=9

14-y=9

-y=-5

y=5

now we know that x=7 and y=5, 2x+y becomes:

2(7)+5

14+5

19
8 0
3 years ago
The following circle graph represents how the Townsends spend their monthly income. What is the central angle measure of the sec
mars1129 [50]
Please say brainliest.

It would be 18 degrees.

5% of 360(the total number of degrees in a circle) = 18 degrees
3 0
3 years ago
FIND THE SUM:<br> (6x^2-2x-3)+(x^2-3x+1)<br><br> 1.6x^2+6x+3<br> 2.7x^2+5x+2<br> 3.7x^2-5x-2
Butoxors [25]

8x^2-5x-2

method

open bracket

6x^2-2x-3+x^2-3x+1

collect like terms

6x^2+2x^2-3-3x+1

8x^2-3-3x+1

again collect like terms

8x^2-2x-3x-3+1

8x^2-5x-3+1. (-×-=+ addition but keeping sign minus

8x-5x-2. (-×+=-)

5 0
2 years ago
Find the volume of a right circular cone that has a height of 2.5 m and a base with a diameter of 18.6 m. Round your answer to t
dem82 [27]

Answer:

226.4 m³.

Step-by-step explanation:

The volume of a right cone with a circular base is:

\displaystyle V = \frac{1}{3} \pi\cdot r^{2}\cdot h,

where

  • V is the volume of the cone,
  • r is the radius of the cone,
  • and h is the height of the cone.

The radius of the circular base is one-half its diameter.

\displaystyle r= \frac{1}{2}h=9.3\;\text{m}.

\displaystyle V = \frac{1}{3} \pi\cdot r^{2}\cdot h = \frac{1}{3} \pi \times 9.3^{2} \times 2.5 \approx 226.4\;\text{m}^{3}.

5 0
3 years ago
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