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satela [25.4K]
3 years ago
14

A student runs 4km in 30 minutes at a steady pace . what is the students average speed in km/hr

Mathematics
2 answers:
Alina [70]3 years ago
6 0
The students run a average speed of 8km/hour 

Talja [164]3 years ago
5 0
If a student runs 4K in 1/2 hours, then he would run 8k In 1 hour, so, 8km/h
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How do i simplify this expression?
levacccp [35]
  First, you try to make sure that both the denominators are the same. In order to do this, you multiply both fractions' numerator and denominator by the other fraction's denominator. Next, you combine like terms in the numerator, and then you simplify the fraction.
8 0
3 years ago
Your favorite dog groomer charges according to your dog's weight. If your dog is 10 pounds or under, the groomer charges $30. If
meriva

Answer:

$170

Step-by-step explanation:

Sally costs 30

Max costs 40

Goliath costs 80(1.25) = 100

add them up

$170

6 0
3 years ago
Calculate the area of triangle ABC with altitude CD, given A (6, -2), B (1, 3), C (5, 5), and D (2, 2).
Flura [38]

Answer:

15 units

Step-by-step explanation:

I just took this geometry test with the same question. Its 15

3 0
3 years ago
Read 2 more answers
Write and equati9n of a l8ne that passes through (1,-2) and is perpendicular to -4x+7y=21
yarga [219]

Given:

A line passes through (1,-2) and is perpendicular to -4x+7y=21.

To find:

The equation of that line.

Solution:

We have, equation of perpendicular line.

-4x+7y=21

Slope of this line is

m_1=-\dfrac{\text{Coefficient of x}}{\text{Coefficient of y}}

m_1=-\dfrac{-4}{7}

m_1=\dfrac{4}{7}

Product of slope of two perpendicular lines is -1.

m_1\times m_2=-1

\dfrac{4}{7}\times m_2=-1

m_2=-\dfrac{7}{4}

Now, slope of required line is -\dfrac{7}{4} and it passes through (1,-2). So, the equation of line is

y-y_1=m(x-x_1)

where, m is slope.

y-(-2)=-\dfrac{7}{4}(x-1)

4(y+2)=-7(x-1)

4y+8=-7x+7

4y+7x=7-8

7x+4y=-1

Therefore, the equation of required line is 7x+4y=-1.

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%20%20%5Cdiv%203x%20-%202%20%3D%201%20%5Cdiv%203x%20-%204" id="TexFormula1" title="x \div 3x
Maurinko [17]

Hello from MrBillDoesMath!

Answer:

x = 1/2 (1 +\- i sqrt(23))

Discussion:

x \3x - 2 =   (x/3)*x - 2   =  (x^2)/3 - 2     (*)

1 \3x - 4   =  (1/3)x - 4                               (**)

(*) = (**)   =>

(x^2)/3 -2 = (1/3)x - 4        => multiply both sides by 3

x^2 - 6 = x - 12                 => subtract x from both sides

x^2 -x -6 = -12                   => add 12 to both sides

x^2-x +6  = 0

Using the quadratic formula gives:

x = 1/2 (1 +\- i sqrt(23))

Thank you,

MrB

4 0
3 years ago
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