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3 years ago

What is the surface area of a sphere with a radius of 14 units?

Mathematics

1 answer:

zloy xaker [14]3 years ago

7 0

Answer:

$SA=784\pi\ units^{2}$

Step-by-step explanation:

we know that

The surface area of a sphere is equal to

$SA=4\pi r^{2}$

we have

$r=14\ units$

substitute

$SA=4\pi (14)^{2}$

$SA=784\pi\ units^{2}$

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Help plz test !! Four more than the product of a number and 10 is at most the opposite of 12.

professor190 [17]

Answer:

-1.6

Step-by-step explanation:

4 + 10x = -12

10x = -12 -4

x= -1.6

5 0

2 years ago

* A Frog sits on a number line at point −10. It is allowed to make up to 4 jumps. First, it can make a jump of 8 units in either

Alika [10]

Answer and explanation:

Assuming the frog jumps from -10 for all jumps, if the frog decides to make all 4 jumps then :

8 units jump= -2 or -18

4 units jump= -6 or -14

2 units jump= -8 or -12

1 unit jump = -9 or -11

If the frog decides to make one jump then skip the others then:

Either it makes 8 units jump= -2 or -18

Or 4 units jump= -6 or -14

Or 2 units jump= -8 or -12

Or 1 unit jump = -9 or -11

If it decides to make 2 jumps and skip the others then:

Either it makes 8 units jump= -2 or -18

and 4 units jump= -6 or -14

Or 2 units jump= -8 or -12

and 1 unit jump = -9 or -11

Or any other mix of units of jump twice

If it decides to make 3 jumps and skip the others then:

Either it makes 8 units jump= -2 or -18

and 4 units jump= -6 or -14

and 2 units jump= -8 or -12

Or 1 unit jump = -9 or -11

and 4 units jump= -6 or -14

and 2 units jump= -8 or -12

Or any other mix of 3 jumps

If it decides to make no jump at all then the frog would remain at -10

8 0

3 years ago

Given the graph of a degree 3 polynomial below, complete the table of values for either the x-value of a zero, or the multiplici

leva [86]

The roots or zeros of a polynomial are the points it crosses the x-axis.

The roots of the polynomial in order from least to greatest are -5, 0 and 5

From the graph (see attachment), we have the following highlights

The graph cross the x-axis at x = -5
The graph cross the x-axis at x = 0
The graph cross the x-axis at x = 5

The points at which the graph crosses the x-axis are the roots of the graph.

So, the roots are: -5, 5 and 5

Read more about the roots of a polynomial at:

brainly.com/question/7921963

8 0

2 years ago

Find the expansion of tan x about the point X = 0.

yan [13]

Answer:

$f(x) = x +\frac{1}{3}x^{3}+....$

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that The Maclaurin series is given by:

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

= 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

= 4sec²(x) tan²(x) + 2sec⁴(x)

= 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) = sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

$f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......$

$f(x) = x +\frac{1}{3}x^{3}+....$

Therefore, the expansion of tan x at x = 0 is

$f(x) = x +\frac{1}{3}x^{3}+....$ .

8 0

3 years ago

On average, it takes Robert 5.32 minutes to drive to school from his house in the mornings, and it takes him 6.01 minutes to dri

Sergio039 [100]

The answer is b

multiply 5.32 times 46 and then 6.01 times 46 and add

or it could be a if you are rounding

4 0

3 years ago