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3 years ago

What is the volume of a cylinder with a height of 15 and a radius of 11 ???????????????

1 answer:

7 0

Volume of a Cylinder = πr²h
V = π * (11)² * 15
v = π * 121 * 15
v = 1815π units³

In short, Your Answer would be Option C

Hope this helps!

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Please answer with steps shown. Thanks!

maks197457 [2]

Answer: q = 40

Step-by-step explanation:

Given the quadratic formula as

x = [-b +/-√(b^2 -4ac)]/2a

b = -14

a = 1

c = q

Difference d between the two roots.

d = [-b + √(b^2 -4ac)]/2a - [-b -√(b^2 -4ac)]/2a

d = 2√(b^2 -4ac)/2a

d = √(b^2 -4ac)/a

And d = 6

Substituting the values of a,b and c. We have;

6 = √[(-14^2) - (4×1×q)]

Square both sides

6^2 = 196 - 4q

4q = 196 - 36

q = 160/4

q = 40

The equation becomes

x^2 - 14x + 40 = 0

3 0

3 years ago

Solve the equation for x.<br> 15 = -2x + 3

bekas [8.4K]

Answer:

Step-by-step explanation:

$-2x+3=15\qquad\text{subtract 3 from both sides}\\\\-2x+3-3=15-3\\\\-2x=12\qquad\text{divide both sides by (-2)}\\\\\dfrac{-2x}{2}=\dfrac{12}{-2}\\\\x=-6$

6 0

3 years ago

Read 2 more answers

A 3cm×10cm rectangle sits inside a circle with radius of 12. Find the area of the shaded region

kolezko [41]

Answer:

422.4cm²

Step-by-step explanation:

area of circle is pi x 12² which is 144 pi

144pi subtract 30 is 422.4

6 0

3 years ago

Find the exact values of sin2 θ for cos θ = 3/18 on the interval 0° ≤ θ ≤ 90°

mote1985 [20]

Answer:

$sin(2\theta)=\frac{\sqrt{35} }{18}$

Step-by-step explanation:

Recall the formula for the sine of the double angle:

$sin(2\theta)=2*sin(\theta)*cos(\theta)$

we know that $cos(\theta)=\frac{3}{18}$ , and that $\theta$ is in the interval between 0 and 90 degrees, where both the functions sine and cosine are non-negative numbers. Based on such, we can find using the Pythagorean trigonometric property that relates sine and cosine of the same angle, what $sin(\theta)$ is:

$cos^2(\theta)+sin^2(\theta)=1\\sin^2(\theta)=1-cos^2(\theta)\\sin(\theta)=\sqrt{1-cos^2(\theta)} \\sin(\theta)=\sqrt{1-(\frac{3}{18} )^2}\\sin(\theta)=\sqrt{1-\frac{9}{324} }\\sin(\theta)=\sqrt{\frac{324-9}{324} }\\sin(\theta)=\sqrt{\frac{315}{324} }\\\\sin(\theta)=\frac{3}{18}\sqrt{35 }$

With this information, we can now complete the value of the sine of the double angle requested:

$sin(2\theta)=2*sin(\theta)*cos(\theta)\\sin(2\theta)=2*\frac{3}{18} \,\sqrt{35} \,\frac{3}{18}\\sin(2\theta)=\frac{2*3*3}{18*18}\,\sqrt{35} \\sin(2\theta)=\frac{\sqrt{35} }{18}$