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3 years ago

A line passes through the points (k+3,-2k) and (4,1) and has a y-intercept of 6. Find the value of k.

1 answer:

4 0

Y = mx + b; b = 6

1 = 4m + 6; m = -5/4 

-5/4 = (1 + 2k) / (4 - k - 3) = (2k + 1) / (1 - k) 

-5(1 - k) = 4(2k + 1) 

5k - 5 = 8k + 4 

-3k = 9 

k = -3

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Hellpp please 20 points for that

Sergio [31]

Answer:

a) 12cm

b) 360cm²

c) 2160cm cubed

Step-by-step explanation:

you would use Pythagoras theorem for A

so half of 18 is 9

15² - 9² = 144

the square root of 144 is 12

the height is 12cm

18 x 20 = 360cm²

csa x L

csa = b x h divide by 2

18 x 12 = 216cm²/2 = 108cm²

108xm² x 20cm = 2160cm cubed

8 0

3 years ago

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6

katrin2010 [14]

Answer:

Part a) The height of the ball after 3 seconds is $49\ ft$

Part b) The maximum height is 66 ft

Part c) The ball hit the ground for t=4 sec

Part d) The domain of the function that makes sense is the interval

[0,4]

Step-by-step explanation:

we have

$h(t)=-16t^{2} +63t+4$

Part a) What is the height of the ball after 3 seconds?

For t=3 sec

Substitute in the function and solve for h

$h(3)=-16(3)^{2} +63(3)+4=49\ ft$

Part b) What is the maximum height of the ball? Round to the nearest foot.

we know that

The maximum height of the ball is the vertex of the quadratic equation

Convert the function into a vertex form

$h(t)=-16t^{2} +63t+4$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$h(t)-4=-16t^{2} +63t$

Factor the leading coefficient

$h(t)-4=-16(t^{2} -(63/16)t)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$h(t)-4-16(63/32)^{2}=-16(t^{2} -(63/16)t+(63/32)^{2})$

$h(t)-(67,600/1,024)=-16(t^{2} -(63/16)t+(63/32)^{2})$

Rewrite as perfect squares

$h(t)-(67,600/1,024)=-16(t-(63/32))^{2}$

$h(t)=-16(t-(63/32))^{2}+(67,600/1,024)$

the vertex is the point (1.97,66.02)

therefore

The maximum height is 66 ft

Part c) When will the ball hit the ground?

we know that

The ball hit the ground when h(t)=0 (the x-intercepts of the function)

$h(t)=-16t^{2} +63t+4$

For h(t)=0

$0=-16t^{2} +63t+4$

using a graphing tool

The solution is t=4 sec

see the attached figure

Part d) What domain makes sense for the function?

The domain of the function that makes sense is the interval

[0,4]

All real numbers greater than or equal to 0 seconds and less than or equal to 4 seconds

Remember that the time can not be a negative number

6 0

3 years ago

What is the surface area of the cube. A. 409 cm² B. 2400 cm² C. 2004 cm² D. 40 cm²

larisa [96]

Side=a=20cm

Surface area

6a^2
6(20)^2
400(6)
2400cm^2

4 0

2 years ago

Help!!!! No Bad Answers Will Give brainiest!!

Alborosie

The domain the given graph is :

-12 < x < 13

6 0

2 years ago

Elimination method - -4x = 4y 4x - 8y = 12

nika2105 [10]

Answer: (4,-4)

y=-4, x=4

Step-by-step explanation:

-4x=4y

4x-8y=12

-8y=-4x+12 Now I can substitute -4x for 4y

-8y=4y+12

-2y=y+12

-3y=12

y=-4

-4x=4y

-4x=4(-4)

x=4

6 0

3 years ago