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mina [271]
2 years ago
9

What is the percent of change from 78 to 100

Mathematics
1 answer:
mojhsa [17]2 years ago
8 0
We have a positive change (increase) of 28.20512821 percent because the new value is greater than the old value.
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What is the slope of a line that contains the points (-8,7) and (-4,-5)?
Pani-rosa [81]

Answer:

-3

Step-by-step explanation:

To find the slope given two points

m = (y2-y1)/(x2-x1)

   = (-5-7)/(-4- -8)

  = (-12)/(-4+8)

  =-12/4

  = -3

7 0
3 years ago
Find the measure of a. A. 110 B. 125 C. 55 D. 75
tangare [24]
The answer to your question is c 55
5 0
2 years ago
Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
3 years ago
5x - (8x -17) = -10
mihalych1998 [28]

Answer:

x=9

Step-by-step explanation:

isolate the variable by dividing each side by factors that dont contain the variables.

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5 0
3 years ago
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Answer:

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Step-by-step explanation:

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