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Otrada [13]
3 years ago
11

What is the approximate area of the shaded sector in the circle shown below?

Mathematics
2 answers:
ozzi3 years ago
7 0

Answer: A) 7.26\ in^2

Step-by-step explanation:

From the given figure, Diameter of circle = 4.3 in.

Radius of the circle = \frac{4.3}{2}\ in.

Central angle = Measure of arc = \theta=180^{\circ}

The area of sector of a circle is given by :-

\text{Area of sector}=\frac{\theta}{360^{\circ}}\times\pi r^2\\\\\Rightarrow\text{Area of sector}=\frac{180^{\circ}}{360^{\circ}}\times(3.14)(\frac{4.3}{2})^2\\\\\Rightarrow\text{Area of sector}=\frac{1}{2}(3.14)(\frac{18.49}{4}\\\\\Rightarrow\text{Area of sector}=7.257325\approx7.26\ in^2

Hence, the approximate area of the shaded sector in the circle  = 7.26\ in^2

lara [203]3 years ago
4 0
A is the answer:)))))))))))))))))))))))
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4/5 minus 1/3 plz i need help
IRISSAK [1]

Answer:

7/15

Step-by-step explanation:

Step 1

We can't subtract two fractions with different denominators. So you need to get a common denominator. To do this, you'll multiply the denominators times each other... but the numerators have to change, too. They get multiplied by the other term's denominator.

So we multiply 4 by 3, and get 12.

Then we multiply 1 by 5, and get 5.

Next we give both terms new denominators -- 5 × 3 = 15.

So now our fractions look like this:

12

15

−  

5

15

Step 2

Since our denominators match, we can subtract the numerators.

12 − 5 = 7

So the answer is:

7

15

Step 3

Last of all, we need to simplify the fraction, if possible. Can it be reduced to a simpler fraction?

To find out, we try dividing it by 2...

Nope! So now we try the next greatest prime number, 3...

Nope! So now we try the next greatest prime number, 5...

Nope! So now we try the next greatest prime number, 7...

Nope! So now we try the next greatest prime number, 11...

No good. 11 is larger than 7. So we're done reducing.

6 0
2 years ago
If 3/7m = 7/3n, then what is the value of m/n ?
Gwar [14]

49/9=m/n this is because you multiply by the reciprocal of 3/7

6 0
3 years ago
What is the solution to this equation x-16=-8
arsen [322]
X-16=-8
Add 16 on both sides 
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And solve; your answer will be
x=8
3 0
2 years ago
Decimal and integers are both classified as ______ numbers
Evgesh-ka [11]

Answer:whole numbers

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.
Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

3 0
3 years ago
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