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ddd [48]
3 years ago
7

You kept track of the numbr of minutes you spent on homework for one week. 18, 20, 22, 11, 19, 18, 18 What is range of minutes y

ou spend doing homework? *
Mathematics
1 answer:
frosja888 [35]3 years ago
4 0

Answer:

your range is 11

Step-by-step explanation:

22-11= 11

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Help, this is a khan question so you may have done it before
Readme [11.4K]

Answer:

Eric, 9.8 meters.

Step-by-step explanation:

Use the distance formula with the given coordinate and each of the other coordinate.

\sqrt{(10-3)^{2}+(8-2)^{2} } Brooke

\sqrt{(-6-3)^{2}+(6-2)^{2} } Eric

\sqrt{(8-3)^{2}+(-6-2)^{2} } Cam

You'll see that Eric kicked the furthest distance with a distance of 9.8 meters.

Brooke kicked 9.2 meters and Cam kicked 9.4 meters. What does the distance formula really mean? When you solve for the distance between two points, you're really just creating a triangle between those two points, and you're actually looking for the hypotenuse. The slope that intersects your two points is the hypotenuse of the triangle you created.

8 0
3 years ago
Let F(x)=x^2-4 and G(x)=5-x<br><br> Find (F/G)(x)=
sammy [17]

Answer:(F/G)(x)=7

Step-by-step explanation:

5 0
3 years ago
In what form is the following linear equation written v=9x+2
Oksi-84 [34.3K]
Taking v to be y, it's written in the standard form for a linear equation, y=mx+c
6 0
3 years ago
A quadrilateral has no pairs of parallel sides.what two shapes could be it
Basile [38]

Answer:

All trapezoids are quadrilaterals.

Step-by-step explanation:

Incorrect. Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides. A trapezoid can never be a parallelogram. The correct answer is that all trapezoids are quadrilaterals.

8 0
3 years ago
Read 2 more answers
A projectile is thrown upward off a 130ft cliff at a velocity of 15ft/second. How long before it reaches a height of 25ft?
zhuklara [117]

Answer:

The time taken for the projectile to reach the given height is 1.2 s.

Step-by-step explanation:

Given;

height of travel, h = 25 ft

initial velocity of the projectile, u = 15 ft/s

The time taken for the projectile to travel a height of 25 ft is given by the following kinematic equation;

h = ut + ¹/₂gt²

25 = 15t + ¹/₂(9.8)t²

25 = 15t + 4.9t²

4.9t² + 15t - 25 = 0

solving the quadratic equation, we will have the following solution of t;

t = 1.2 s

Therefore, the time taken for the projectile to reach the given height is 1.2 s.

5 0
3 years ago
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