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2 years ago

You run 1 3\4 miles each day. Your friend runs 4 times as far as you do. How far does your friend run each day ? Please help

Mathematics

1 answer:

solniwko [45]2 years ago

7 0

Well to find the answer multiply the miles you run each day by 4 which is 7. the friend runs 7 miles each day

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there were three apples , 1 mixed fruit, 2 grape, and 4 tropical fruit juice boxes in the cooler at the picnic. What is the prob

ivanzaharov [21]

Answer:

1/45 ≈ 0.022

Step-by-step explanation:

There are originally 10 juice boxes in the cooler, 2 of which are grape.

The probability that the first juice box is grape is 2/10.

The probability that the second juice box is grape is 1/9.

The probability of both is (2/10) (1/9) = 1/45 ≈ 0.022.

3 0

3 years ago

Por favor denme una ecuación a esta pregunta:

Katena32 [7]

Answer:

x + 13 = 25

Step-by-step explanation:

Se puede representar "un número" con x.

"Adiciona" significa +.

x + 13

"El resultado" significa =.

x + 13 = 25

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3 years ago

Solve for x. Enter the solutions from least to<br> greatest.<br> .<br> (x - 10)2 - 1 = 0

lawyer [7]

2x -20 -1 =0
2x -21=0
2x=21
x= 10.5

4 0

2 years ago

Find the product <br> (4p + 6)( 7p -5)

sleet_krkn [62]

Answer:

28p^2+22p-30

Step-by-step explanation:

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3 0

2 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

4 0

2 years ago