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3 years ago

Find what the question asks. 200 points. Will mark brainleist

Mathematics

2 answers:

kotegsom [21]3 years ago

6 0

Note:
The area of a circle with radius r is πr².
An entire circle has a central angle of 360°.

For the sector with radius = (3+4) = 7cm and a central angle of 120°, the area is
$A_{1}= \frac{120}{360}*(49 \pi ) = \frac{49}{3} \pi \, cm^{2}$

For the sector with radius = 3 cm and a central angle of 120°. the area is
$A_{2}= \frac{120}{360} *(9 \pi ) = 3 \pi \, cm^{2}$

The area of the shaded sector (see the figure) is
$A=A_{1}-A_{2} = ( \frac{49}{3} -3) \pi = \frac{40}{3} \pi \, cm^{2}$

Answer: $\frac{40}{3} \pi \, cm^{2}$

WARRIOR [948]3 years ago

6 0

Answer:

hey i have the same question.. do u have the answer yet

Step-by-step explanation:

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2 years ago

One edge of a cube is measured and found to be 13 cm. the volume of the cube in m3 is ________.

slavikrds [6]

Since a cube is defined to have the same measurement in all three dimensions, it's volume can be calculated by raising the measurement of one side to the power of three.

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3 years ago

Need to find shaded area of figure please help!

devlian [24]

Answer:

39 ft^2

Step-by-step explanation:

Find the area of the triangle and subtract the area of the square.

area of triangle = bh/2

area of square = s^2

shaded area = bh/2 - s^2

shaded area = (20 ft)(16 ft)/2 - (11 ft)^2

shaded area = 160 ft^2 - 121 ft^2

shaded area = 39 ft^2

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3 years ago

What is the value of x?

gtnhenbr [62]

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2 years ago

Read 2 more answers

Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

jok3333 [9.3K]

Answer:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

$\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}$

And, the distance from P to (0, -2) will be:

$\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}$

So sum of the two distances must be 6. Therefore:

$d_1+d_2=6$

Now, by substitution:

$(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6$

Simplify. We can subtract the second term from the left:

$\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$

Square both sides:

$(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)$

We can cancel the x² terms and continue squaring:

$y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4$

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

$-8y=36-12\sqrt{x^2+(y+2)^2}$

We can divide both sides by -4:

$2y=-9+3\sqrt{x^2+(y+2)^2}$

Adding 9 to both sides yields:

$2y+9=3\sqrt{x^2+(y+2)^2}$

And, we will square both sides one final time.

$4y^2+36y+81=9(x^2+(y^2+4y+4))$

Distribute:

$4y^2+36y+81=9x^2+9y^2+36y+36$

The 36y will cancel. So:

$4y^2+81=9x^2+9y^2+36$

Subtracting 4y² and 36 from both sides yields:

$9x^2+5y^2=45$

And dividing both sides by 45 produces:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

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3 years ago

Read 2 more answers