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3 years ago

7

The long jump distance of four students are listed in the table below. How much farther did the student with the longest distanc

e jump than the student with the shortest distance.
Joe 6.25
Shakira 4.37
Jackson 5.85
Chelsey 4.20

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

5.85

Step-by-step explanation:

how long she is gonna be jumping

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Need help can someone help me pleaseeee

Hunter-Best [27]

at first we write each one as a sequence

and find its equation

the first table

-19, -11, -3, 5

8x-19

the third table

15,12,9,6

-3x+15

the second table

-1. 5,1.5,3,4.5

+3 +1.5 +1.5

so the third table is the nonlinear function

8 0

3 years ago

An in ground rectangular pool has a concrete pathway surrounding the pool. if the pool is 16 feet by 32 feet and the entire area

Anastasy [175]

1. Let's call:

x: the width of the walkway.
b: 2x+16 (there is a widht "x" of the walkway on each side).
h: 2x+32 (there is a widht "x" of the walkway on each side).
A: 924 ft^2 ( the area of the pool of including the walkway).

2. The area of a rectangle is:

A=(b)(h)

b: the base of the rectangle (2x+16)
a: the height of the rectangle (2x+32)

3. Then, we have a quadratic equation:

924=(2x+16)(2x+32)
4x^2+64x+32x+512-924=0
4x^2+96x-412=0

4. We apply the quadratic formula to find the value of "x":

x=(-b±√(b^2-4ac))/2a

x=3.71

The answer is: The width of the walkway is 3.71 ft

7 0

3 years ago

Simplify (3x2-5x-6) - (5x2+4x+4)

Lisa [10]

Answer:

=−2x^2−9x−10

Step-by-step explanation:

would you like an explantion or are you good

8 0

2 years ago

In the picture below, ABCD is a rectangle. If the area of triangle ABE is 40, what is the area of the triangle?

MariettaO [177]

Area of the rectangle: 112

Step-by-step explanation:

Picture is missing: find it in attachment.

The area of a triangle is given by

$A=\frac{1}{2}bh$

where

b is the length of the base

h is the height

For triangle ABE, we know that

A = 40 (area)

h = 8 (height)

So we can find the length of the base AE:

$AE=b=\frac{2A}{h}=\frac{2(40)}{8}=10$

Now we observe that the base of the rectangle, AD, is the sum of AE and DE, therefore:

$AD=AE+DE=10+4=14$

We also know the height of the rectangle AB is 8, and that the area of a rectangle is

$A=bh$

where b is the base and h the height. Therefore, the area of this rectangle is

$A=bh=(AD)(AB)=(14)(8)=112$

Learn more about area of figures:

brainly.com/question/4599754

brainly.com/question/3456442

brainly.com/question/6564657

#LearnwithBrainly

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3 years ago

Particle 1 of charge q1 �� 5.00q and particle 2 of charge q2 �� 2.00q are fixed to an x axis. (a) as a multiple of distance

Naya [18.7K]

<span>Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get </span>

<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>

Rearranging to calculate for d:

<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207 </span>

<span>
We pick the value that is > q2 hence,</span>

d = 2.72075922005613*L

<span>d = 2.72*L</span>

3 0

2 years ago