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strojnjashka [21]
3 years ago
9

Make t the subject of this formula. r = 7 (t + 3)

Mathematics
1 answer:
kari74 [83]3 years ago
3 0

Answer:

t= <u>r-21</u>

7

Step-by-step explanation:

r = 7(t+3)

expanding the bracket

r = 7t + 21

bringing t to the left hand side

7t = r - 21

divide both sides by seven

t= <u>r-21</u>

7

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2m + 7m + 3m = 72<br><br> whats the answer step by step
Korolek [52]

Answer:

m = 6

Step-by-step explanation:

First, add like terms:

2m + 7m + 3m = 72

12m = 72

Divide each side by 12:

m = 6

So, the answer is m = 6

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Solve the following system of equations using substitution <br> x = 4 <br> 2x + 4y = 36
DiKsa [7]
Just plug in 4 for the x

2*4 +4y =36
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What is 1+1 = plz anser me
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How do you divide 4/3 by 1.5
bixtya [17]

Answer:

0.9

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Find the area of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximati
Y_Kistochka [10]

Hello from MrBillDoesMath!

Answer:   8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..


Area of semicircle = 1/2 * Pi * R^2        

    Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

     = 4 so the semicircle area is

    (1/2) * Pi * (4^2) = (1/2) * Pi * 16 = 8 Pi


Area of triangle.

   First of all, angle ACB is a right angle ( i.e. 90 degrees).

     * This is the Theorem of Thales from elementary Plane Geometry. *

  so by Pythagoras

    AC^2 + BC^2 = AB^2

 But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

 Substituting these in Pythagoras gives

    AC^2 + 4^2 = 8^2 or

    AC^2 = 8^2 - 4^2- = 64 - 16 = 48

    Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

   (1/2) b * h = (1/2)  BC * AC = (1/2) 4 * (4 * sqrt(3)) =  8 sqrt(3)

We conclude the area area of the shaded part is

  8 PI - 8 sqrt(3)   = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx  1.7 so (PI - sqrt(3)) is a positive number, as it better well be!


Thank you,

MrB

6 0
3 years ago
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