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Anastasy [175]
3 years ago
10

Consider the quadratic function that has x-intercepts of -1 and 7 and passes through the point (-2,-20). What is the

Mathematics
1 answer:
Alexxandr [17]3 years ago
4 0

Answer:

a = -20/9.

Step-by-step explanation:

We can write it as

f(x) = a(x - 7)(x + 1)

When x = -2 and y = -20   (at the point(-2, -20):

-20 = a(-2 -7)(-2 + 1)

-20 = -9*-1 a

9a = -20

a = -20/9.

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Help on this question
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Step-by-step explanation:

always write what you know

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6 0
3 years ago
How much time required to complete a10days work if it could complete by 3 worker in 4days and 6worker in 3.05 days,if 5 workers
Andrei [34K]

Answer:

This seems to be a linear relationship:

A linear relationship can be written as:

y = a*x + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

We know that 3 workers can complete the job in  4 days.

Then if we define W as the rate at which a single worker can complete the work, we have the ordered pair (3 workes, 4 days)

And we also know that 6 workers can finish the job in 3.05 days, then we also have the ordered pair (6 workers, 3.05 days)

Then using these two, we can find the slope of the line:

a = (3.05 days - 4 days)/(6 workers - 3 workers) = -0.35 days/worker.

Then we have the equation:

y = (-0.35 days/worker)*x + b

Where:

y represents the time to complete the job

x represents the number of workers.

b is the y-intercept, to find the value, we could just replace one of the points in the equation, for example with the point (3 workes, 4 days) we need to replace x by 3 workers, and y by 4 days:

4 days = (-0.35 days/worker)*3 workers + b

4 days = -1.4 days + b

4 days + 1.4 days = 5.4 days  = b

then the equation is:

y = (-0.35 days/workers)*x + 5.4 days.

Now we can find the number of days needed for 5 workers to finish the job, we just need to replace x by 5 workers:

y = (-0.35 days/workers)*5 workers + 5.4 days

y = 3.65 days.

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3 years ago
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