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Bad White [126]

3 years ago

10

What is the arc length of PQ?

1 answer:

almond37 [142]3 years ago

3 0

Answer:

43.9823

Step-by-step explanation:

$Arc length = 2\pi r(\frac{0}{360} )\\ = 2\pi 12(\frac{210}{360} )$

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Help please and simply it

Elena-2011 [213]

-6, because -6^3 is -215

5 0

3 years ago

Juliette made cookies for a bake sale and wants to package the same number of cookies together. If she makes 16 chocolate chip c

pentagon [3]

Answer:

8 cookies

Step-by-step explanation:

We are asked in the above question to find the most number of cookies that would be in the bag.

We solve the above question using the Greatest Common Factor method

We find the factors of 16, 24, 32

The factors of 16 are: 1, 2, 4, 8, 16

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

The factors of 32 are: 1, 2, 4, 8, 16, 32

Then the greatest common factor is 8.

Therefore, the most number of each type of cookie that will be in the bag is 8 cookies

4 0

3 years ago

Pl help it’s for a grade

zlopas [31]

Answer:

width = x+2

Step-by-step explanation:

area/length = width

(x^2+9x+14) / (x+7) =?

(x+7)(x+2) / (x+7) = x+2

5 0

3 years ago

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

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

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

<em>Hence, circle Q has a smaller radius</em>

4 0

3 years ago

Please Help!!!!! <br>If secθ = 4, then sinθ = _____.

KiRa [710]

Knowing that $\sec x=\dfrac{1}{\cos x}$ :

$\sec\theta=4\iff\dfrac{1}{\cos\theta}=4\iff\cos\theta=\dfrac{1}{4}$

Using that $\sin^2x+\cos^2x=1$ :

$\sin^2\theta+(\dfrac{1}{4})^2=1\iff \sin^2\theta=\dfrac{15}{16}\iff\\\\\boxed{\sin\theta=\pm\dfrac{\sqrt{15}}{4}}$

8 0

3 years ago