All categories

4 years ago

Paulie and Joanne ran 10km in all. Joanne ran 400 meters farther than Paulie. How far did each of them run?

Mathematics

2 answers:

Ira Lisetskai [31]4 years ago

8 0

They at least ean 2 miles each im not really sure

musickatia [10]4 years ago

3 0

Joanne ran 0.4, so Paulie must have ran 9.6

You might be interested in

Type integer or fraction as answer simplify answer

vova2212 [387]

Answer:

51/28

Step-by-step explanation:

Start by making common denominators.

7 2

4 4

Now subtract the two.

7 - 2 = 5

Now you have to make common denominators for the second part, 4/7

16 + 35 = 51

28 28 28

brainliest is nice ig

7 0

3 years ago

What is the image of (-8,6) after a translation over the line y=-x

stira [4]

Answer:

(-6, 8)

Step-by-step explanation:

(x, y) --> (-y, -x)

(-8, 6) --> (-6, 8)

Therefore, (-6, 8)

I hope this helped and have a good rest of your day!

8 0

3 years ago

In 3-way light bulbs, the highest wattage is the sum of the two lower ones. If the lowest is 60 watts, and the highest 150 watts

otez555 [7]

Since 150= 60 +x
x=110

3 0

3 years ago

Given points A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1), which of the following proves that

damaskus [11]

Similar triangles may or may not have equal side lengths

Triangles ABC and DEF are similar by SSS theorem

<h3>How to determine the similarity statement</h3>

The coordinates of the two triangles are:

A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1)

Calculate the lengths of both triangles using the following distance formula

$d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}$

For triangle ABC, we have:

$AB = \sqrt{(0 -4)^2 + (0 -0)^2} = 4$

$BC = \sqrt{(4 -0)^2 + (0 -2)^2} = 2\sqrt 5$

$CA = \sqrt{(0 -0)^2 + (0 -2)^2} = 2$

For triangle DEF, we have:

$DE = \sqrt{(0 -2)^2 + (0 -0)^2} = 2$

$EF = \sqrt{(2 -0)^2 + (0 -1)^2} = \sqrt 5$

$FD = \sqrt{(0 -0)^2 + (1 -0)^2} = 1$

Divide corresponding sides of the triangles to calculate the scale factor k

$k = \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$

This gives

$k = \frac{4}{2} = \frac{2\sqrt 5}{\sqrt 5} = \frac{2}{1}$

Evaluate the quotients

$k = 2 = 2 = 2$

Since the quotients are equal, then the triangles are similar

Hence, triangles ABC and DEF are similar by SSS theorem

Read more about similar triangles at:

brainly.com/question/14285697

3 0

2 years ago

Can some one help me with this？

pychu [463]

Answer:

8 in

Step-by-step explanation:

Area = (diagonal 1 x diagonal 2)/2

60 = (diagonal 1 x 15)/2

60 x 2 = diagonal 1 x 15

120 = diagonal 1 x 15

diaognal 1 = 120/15 = 8

3 0

3 years ago