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Liula [17]
3 years ago
11

The person who answers the question with the mathematical explanation and answer gets 65 points in total if answers correctly an

d marked as the brainliest so DON'T ANSWER WITH AN IDK AND GET FREE POINTS! If the area of a circle of radius 7 cm is 154 sq cm, what is the area of the shaded portion of the figure shown below?

Mathematics
2 answers:
liubo4ka [24]3 years ago
8 0

Answer:

10.5 squared centimetres.

Step-by-step explanation:

First, find the area of that sector. We are already given that the circle is 154 squared centimetres.

We are given 90 degrees (or one fourth of the circle). Thus, divide 154 by 4 to get the area of the sector (the white area).

The area of the sector is 154/4 or 38.5

We then need to find the area of the square. After that, we can subtract the sector area (the white area) from the area of the square (the total area) to find the area of the shaded section.

The area of the square is 7*7 or 49.

Thus, the shaded portion will be 49-38.5 or 10.5 squared centimetres.

sweet [91]3 years ago
5 0

Answer:

10.5 cm^2

Step-by-step explanation:

The square has area s^2 = (7 cm)^2 = 49 cm^2

The area of the portion is the the area of the square minus the area of 1/4 of the circle.

Area of 1/4 of the circle = 154 cm^2 / 4 = 38.5 cm^2

Area of shaded portion = 49 cm^2 - 38.5 cn^2 = 10.5 cm^2

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6 0
3 years ago
Let f(x)=7x^2-8. find f(a^2)
strojnjashka [21]

Answer:

\large\boxed{f(a^2)=7a^4-8}

Step-by-step explanation:

f(x)=7x^2-8\\\\f(a^2)\to \text{put}\ x=a^2\ \text{to the equation of the function}\ f:\\\\f(a^2)=7(a^2)^2-8\qquad\text{use}\ (a^n)^m=a^{nm}\\\\f(a^2)=7a^{2\cdot2}-8\\\\f(a^2)=7a^4-8

6 0
3 years ago
1. Two object accumulated a charge of
In-s [12.5K]

Answer:

The  magnitude of the electric force  between these two objects

will be: 181.274 N.

i.e.  F  =181.274 N

Step-by-step explanation:

As

Two object accumulated a charge of  4.5 μC and another a charge of 2.8  μC.

so

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8  μC = 2.8 × 10⁻⁶  C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}

   =\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}

   =\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}        ∵ 4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6

   =\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}    ∵ \mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4

\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}

  =\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}

  =\frac{12.6}{0.025^2\times \:35.4\pi }        ∵ \left(10^{-2}\times \:2.5\right)^2=0.025^2

  =\frac{12.6}{0.022125\pi }        ∵ 35.4\pi\times 0.025^2=0.022125\pi

   =\frac{12.6}{0.06950 }

F  =181.274 N

Therefore, the  magnitude of the electric force  between these two objects will be: 181.274 N.

i.e.  F  =181.274 N

7 0
3 years ago
X + 8x + 15 help me out with this plz
aleksklad [387]

Answer:

3(3x + 5)

Step-by-step explanation:

x + 8x + 15              Combine liked terms

9x + 15

GCF: 3

3(3x +5)                   Factor

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3 years ago
What is the value of the expression 3a+b2 when a=14 and b=32 ?
soldi70 [24.7K]

Answer:

106

Explanation:

3a+b2

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3(14) + 2(32)

Solve:

3(14) + 2(32)

42 + 64

106

5 0
3 years ago
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