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DIA [1.3K]
4 years ago
13

What is the simplified form of (2 X 102)2 in standard notation? Do not use commas in your answer. (HAS TO BE A NUMBER)

Mathematics
1 answer:
Oxana [17]4 years ago
7 0

Answer:

40000

Step-by-step explanation:

(2 \times  {10}^{2} )^{2}  \\  =  {2}^{2}  \times  {10}^{2 \times 2} \\  = 4 \times  {10}^{4}  \\  = 40000

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The students in Mr. Wilson's Physics class are making golf ball catapults. The
Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

x=\frac{-0.68(+)0.68} {(-0.028)}=0

x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A  reasonable domain is the distance between the two x-intercepts

so

0 \leq x \leq 48.6\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

0 \leq y \leq 8.3\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0
3 years ago
The square pyramid represented below has a base edge of 6 inches and a height of 5 inches. What is the volume in cubic inches of
lawyer [7]
6*6*5*(1/3)=60in cubed
8 0
3 years ago
Read 2 more answers
Find the missing side and angles.
sweet [91]

Answer:

Step-by-step explanation:

This is an equilateral triangle.

The altitude bisects the 60 degree angle, so a=30

b=60 since its an interior angle of the isosceles triangle.

e = 4 since the two triangles formed by the altitude are congruent.

c = 8 since the side length is 4 + 4 = .

f = 4sqrt3 by the properties of 30 60 90 triangles.

5 0
2 years ago
Solve this inequality: 3b – 7 &lt; 32
butalik [34]
It's a lot like solving equations.

3b-7 < 32
Add 7 to each side.
3b < 39
Divide by 3 on each side.
b < 13

Hope this helps :)
6 0
3 years ago
Read 2 more answers
g The probability that a university graduate will be offered no jobs within a month of graduation is estimated to be 5%. The pro
dexar [7]

Answer:

49% probability that a graduate is offered fewer than two jobs

Step-by-step explanation:

We have these following probabilities:

5% probabilities of not being offered a job

44% probability of receiving one job offer

28% probability of received two job offers.

23% probability of receiving three job offers.

Determine the following probabilities: A. P(A graduate is offered fewer than two jobs)

Zero or one

5 + 44 = 49%

49% probability that a graduate is offered fewer than two jobs

8 0
3 years ago
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