Ask question

Ask question

All categories

3 years ago

7

If the sphere shown above has a radius of 5 units, then what is the approximate volume of the sphere? (Use 3.14 for .) A. 261.67

cubic units B. 523.33 cubic units C. 314 cubic units D. 654.17 cubic units

1 answer:

Olenka [21]3 years ago

8 0

Answer:

B. 523.33 cubic units

Step-by-step explanation:

The volume of a sphere is given by

V = 4/3 pi r^3

V = 4/3 (3.14) 5^3

V =523.3333333

V = 523.33 units^3

You might be interested in

Kerry’s grandmother, Mrs. Reynoso, is helping Kerry pay for guitar lessons. She has set up a special savings account to pay for

s2008m [1.1K]

Answer:

Month 5 is when it will be at $100

I hope this helps!

Step-by-step explanation:

x---0-----1-----2-----3----4-----5

y-300-260-220-180-140-100

y=40x+300

5 0

3 years ago

Find the equation of a line that passes through the point (0,1) and has a gradient of 2.

Usimov [2.4K]

Answer:

y = 2x + 1

Step-by-step explanation:

Given point (0, 1) and the slope m = 2

<u>Use point- slope form to find the equation of the line:</u>

y - y₁ = m(x - x₁)
y - 1 = 2(x - 0)
y - 1 = 2x
y = 2x + 1

6 0

2 years ago

What’s the area of this figure?

Ivan

Answer:

C. 155.5 cm2

Step-by-step explanation:

area = 5*15 + (15+8)*7/2 = 75 + 80.5 = 155.5 cm2

3 0

3 years ago

How do u simplify this expression |-7|

snow_tiger [21]

The absolute value is 7.

7 0

3 years ago

Read 2 more answers

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago