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sergiy2304 [10]
3 years ago
11

(-4, -2),(-18,7) slope

Mathematics
1 answer:
Nuetrik [128]3 years ago
3 0

Slope = (7 + 2)/(-18 + 4) = 9/-14= - 9/14

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Find the value of x
Novay_Z [31]

Answer:

<em>x=3</em>

Step-by-step explanation:

<u>Angles and Lines</u>

We must recall:

  • The sum of the internal angles of a triangle is 180°
  • Two linear angles add up to 180°

The image shows a triangle and the extension of its base to form two linear angles.

Note from the triangle, angle R is what it takes to get to 180°:

m\angle R=180-25x-(57+x)

From the line SQ, angle R what it takes to get to 180°. Thus:

m\angle R=180-45x

Equating:

180-25x-(57+x)=180-45x

Subtracting 180 and simplifying:

-25x-57-x=-45x

19x=57

Dividing by 19:

x=3

4 0
2 years ago
In an ore, 9.8% of its total weight is metal. How many pounds of metal are in 1,950 lb of ore?
Fudgin [204]

Answer

Find out the  how many pounds of metal are in 1,950 lb of ore .

To proof

let us assume that the pounds of metal are in 1,950 lb of ore be x .

As given

In an ore, 9.8% of its total weight is metal.

ore weight = 1,950 lb

9.8% is written in the decimal form

= \frac{9.8}{100}

= 0.098

Than the equation becomes

x = 0.098 × 1950

x = 191.1 pounds

Therefore the 191.1 pounds of metal are in 1,950 lb of ore .

Hence proved



5 0
3 years ago
The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time
cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

1\leq t\geq \sqrt{2}

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

h =20-5t^{2}-----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height h = 15m  

Put height value in equation 1

15 =20-5t^{2}

5t^{2} =20-15

5t^{2} =5

t^{2} =1

t =\pm1

We know that the time is always positive, therefore t=1

when the stuntman is 10m above the ground.

height h = 10m  

Put height value in equation 1

10 =20-5t^{2}

5t^{2} =20-10

5t^{2} =10

t^{2} =\frac{10}{5}

t^{2} =2

t=\pm\sqrt{2}

t=\sqrt{2}

Therefore ,time interval of camera film him is 1\leq t\geq \sqrt{2}

7 0
3 years ago
Please help me ASAP only answer if you’re absolutely sure and/or you took the benchmark
ICE Princess25 [194]

Answer: the answer is a

Step-by-step explanation:

8 0
3 years ago
How do you solve?3(t+4)-2(2t+3)=-4
shepuryov [24]
3t+12-4t-6=-4
-t=-4+6-12
-t=-10
t=10
That's your answer.
8 0
3 years ago
Read 2 more answers
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