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Ymorist [56]
3 years ago
5

What is the slope-intercept form of the equation of a line that passes through (5, -4) and has a slope of 3/4?

Mathematics
2 answers:
olasank [31]3 years ago
6 0
The equation of the line that passes through the point (5, -4) and a slope of 3/4 is y =  \frac{3}{4} x -  \frac{31}{4}


garik1379 [7]3 years ago
4 0
First write the equation in point-slope form.

point slope : y - y1 = m (x - x1)

y - (-4) = 3/4 (x - 5)

Now simplify:

y + 4 = 3/4 (x - 5)
y + 4 = 3/4x - 3.75
y = 3/4 - 7.75

Hope this helps :)
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Solve for xxx in each of the figures below and mark the correct value.
Brums [2.3K]

Answer:

Picture 1:

12x+1=109

12x=108

x=9

Picture 2:

2x-3=57

2x=60

x=30

Let me know if this helps!

7 0
3 years ago
A math test is worth 100 points and has 38 problems. each problem is worth either 5 points or 2 points. how many problems of eac
blondinia [14]
Hey there,
There are 2 ways
1st way:
Since there are 8 5 point problems, 38 - 8 = 30, will give you the 2 point problems.

2nd way: 
8 x 5 = 40
100 - 40 = 60
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Thus there are 30, 2 point problems.

Hope this helps :))

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4 0
3 years ago
Read 2 more answers
Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that P = 330

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

A(t) = Pe^{rt}

590 = 330e^{8r}

e^{8r} = \frac{590}{330}

e^{8r} = 1.79

Applying ln to both sides:

\ln{e^{8r}} = \ln{1.79}

8r = \ln{1.79}

r = \frac{\ln{1.79}}{8}

r = 0.0726

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

A(t) = 330e^{0.0726t}

A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

4 0
3 years ago
F(x)=−x <br> 2<br> −7x Find f(−10)
djyliett [7]

Answer:

What's '2−7x'

Step-by-step explanation:

3 0
2 years ago
I need the answer to all of these with work shown. Thanks!!
professor190 [17]

This is for question 3a.

6 0
3 years ago
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