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Lostsunrise [7]
2 years ago
7

Using algebra, find the point at which the line k(x) = 5x - 1 intersects with the line h(x) = -3x - 1 ?

Mathematics
1 answer:
vichka [17]2 years ago
8 0

The point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Given: k(x) = 5x - 1, h(x) = -3x - 1

We need to find the point(if any) at which these two lines k and h meets.

To find point of intersection(if any), we need to set the functions equal as at the point of intersection the (x, y) value will be same for both of the lines.

Therefore, k(x) = h(x)

=> 5x - 1 = -3x - 1

=> 8x = 0

=> x = 0

k(x=0) = 5 * 0  - 1 = -1

Hence the point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Know more about "point of intersection" problems here: brainly.com/question/16929168

#SPJ1

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Jack plotted the graph below to show the relationship between the temperature of his city and the number of ice cream cones he s
timurjin [86]

Answer:

(a) The amount of ice cream increase as temperature increases

(b)

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m = 2 --- slope

Step-by-step explanation:

Given

See attachment for graph

Solving (a): The relationship between the variables

From the attached graph, the dots on the graph increases towards up-right direction. This implies that there is a positive correlation between the variables.

In other words;

The amount of ice cream increase as temperature increases

Solving (b): The line of best fit

First, we draw a line through the points (the line should have almost equal points on both sides; see attachment 2).

From (2), we select any 2 points on the line:

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The slope (m) is:

m = \frac{y_2 -y_1}{x_2 - x_1}

m = \frac{5-35}{0-15}

m = \frac{-30}{-15}

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y = m(x - x_1) + y_1

Substitute known values:

y = 2(x - 0) + 5

y = 2(x) + 5

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6 0
2 years ago
A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet
Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

L_{1} = 2W_{1}

Also, the area of the walkway of 2 feet wide is 448:

W_{2} = 2 ft

A_{T} = W_{2}*L_{2} = 448 ft^{2}

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

A_{T} = A_{1} + A_{w}    (1)          

The area of the pool is given by:

A_{1} = L_{1}*W_{1}        

A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}  (2)          

And the area of the walkway is:

A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}    (3)          

Where the length of the bigger rectangle is related to the lower rectangle as follows:                  

L_{2} = 4 + L_{1} = 4 + 2W_{1}   (4)        

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A_{T} = A_{1} + A_{w}

A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}                

448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}            

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224 = W_{1}^{2} + 8 + 6W_{1}

By solving the above quadratic equation we have:

W₁ = 12 ft

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L_{1} = 2W_{1} = 2*12 ft = 24 ft

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!                                                                                          

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Step-by-step explanation:

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