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Furkat [3]
2 years ago
12

How do i find the measure of these? i’m lost in math

Mathematics
1 answer:
Rina8888 [55]2 years ago
7 0

Answer: Answers and work are all in the picture that I attached. Hope this helps

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What are some equivalent fractions for the problem :<br> 9/2 ?<br><br> Thank you for your time :)
Varvara68 [4.7K]
First, if you convert the improper fraction 9/2 to a mixed number, it will be equal to 9/2.

9/2 = 4 1/2

And if you multiply the top and bottom of a fraction with an integer number, you will have a fraction that is equal to fraction you began with.

I'll use 5 for this example.

9/2 * 5/5 = 45/10
5 0
3 years ago
what is the height of a rectangular prism that has a volume of 2880 cubic inches and a base of 240 inches
Sergeu [11.5K]

Answer:

h = 12 in

Step-by-step explanation:

To find the height, divide the volume by the base:

              2880 in^3

height:  ------------------ = 12 in

                  240 in

4 0
2 years ago
Read 2 more answers
If the geometric mean of a and 28 is 16 sqrt 7 find a
kozerog [31]

Answer:

a = 64

Step-by-step explanation:

The geometric mean of 2 numbers a and b is \sqrt{ab} , then

\sqrt{28a} = 16\sqrt{7} ( square both sides )

28a = (16\sqrt{7} )² = 1792 ( divide both sides by 28 )

a = 64

8 0
2 years ago
Read 2 more answers
19 times 15 with estimating
VladimirAG [237]
284....................
3 0
3 years ago
Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt
Leni [432]
Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = \sqrt{Δx^2+Δy^2}
d₁ = \sqrt{(-2-1)^2+(1-4)^2}
d₂ = \sqrt{(x-1)^2+(y-4)^2}
d₁ = d₂
∴ \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
 (x-1)²+(y-4)² = 18
from equatoin (1)  y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = \pm \sqrt{9} = \pm 3
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1)  or  (-2,7)












8 0
3 years ago
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