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nata0808 [166]
3 years ago
5

The point (0,-5) lies ln a straight line with gradient = 1/5. Find the equation of the line

Mathematics
1 answer:
Zolol [24]3 years ago
5 0

Answer:

y = 1/5x - 5.

Step-by-step explanation:

Use point-slope form of a straight line:

y - y1 = m(x - x1)

Here m = the gradient = 1/5 and (x1, y1) = (0, -5) so we have:

y - -5 = 1/5(x - 0)

y + 5 = 1/5x

y = 1/5x - 5.

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Kevin can mow the lawn in 1.5 hours. Together, Kevin and Eric can mow the lawn in 30 minutes. How long will it take Eric to mow
xeze [42]

Answer:

45

Step-by-step explanation:

These work problems are always done in terms of fractions ie if both (Kevin & Eric) took 30mins to mow and Kevin took 1.5hrs (90mins) alone then we can make an equation like below

1/time took for both = 1/time took for Kevin + 1/time took for Eric

1/30 = 1/90+1/x    ==> calling time took for Kevin x

solving the above 1/x = 1/30-1/90

1/x=2/90

x=45

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3 years ago
Write an equation in point-slope form of the line that passes through (-4,1) and (4,3).
SVEN [57.7K]

Step-by-step explanation:

y-y₁=m(x-x₁) 

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5=m

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3 years ago
Calculate the perimeter of a regular pentagon which has a side of 9cm
Alisiya [41]
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hope this helps.
8 0
3 years ago
Distributive property to express 16+32
azamat
4 times 4 plus 9 times 4
5 0
4 years ago
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Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water in
nadezda [96]

Answer:

11/4 seconds

Step-by-step explanation:

We can see from the given equation that the height at the diving board is h = 3.

This is because the h(t) equation has that +3 at the end, which denotes the initial height of the diver, the height when he is standing on the board before jumping.

To find where h(t) = 3 is true, we need to set h(t) equal to 3 and solve for t.

h(t) = 3

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so h(t) = 3 when t = 0 and when t = 11/4 sec

we already know that at t = 0 the height is 3, it is the initial height given from the equation, so we want to use the other solution for t.

the diver is back at the height of the diving board at t = 11/4 sec

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3 years ago
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