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Dafna11 [192]
3 years ago
13

Pls i need this, it’s due today. i will mark brainliest.

Mathematics
1 answer:
MArishka [77]3 years ago
8 0

Answer:

Step-by-step explanation:

Y intercepts : 5,-1,3,7,1

X- intercepts: 2,3,1,-1

Its decreasing

It is a negative slope

Maximum : 7 for y

Min : -1

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The market is located at point (−4, −3) on the coordinate graph. The barber shop is 8 units to the right and 4 units up from the
Evgesh-ka [11]

Answer:

(4, 1)

Step-by-step explanation:

8 units to the right = x + 8, so -4 + 8 = 4

4 units up = y + 4, so -3 + 4 = 1

4 0
3 years ago
3+-√(-3)^2 - 4(5)(-1)
Alona [7]

Answer:

Step-by-step explanation:

Easy way to do this is step by step.  Your quadratic, from your entry, must be

5x^2-3x-1.

Step by step looks like this, one thing at a time:

x=\frac{3+\sqrt{(-3)^2-4(5)(-1)} }{2(5)} becomes

x=\frac{3+\sqrt{9-(-20)} }{10} becomes

x=\frac{3+\sqrt{9+20} }{10}

and this of course is

x=\frac{3+\sqrt{29} }{10}

Do the same with the subtraction sign to get the other solution.

If you're unsure of how to enter it into your calculator, do it step by step so you don't mess up the sign.  If you enter it incorrectly, you could end up with an imaginary number when it should be real, or a real one that should be imaginary.

Just my advice as a high school math teacher.

3 0
3 years ago
Help please I’ll mark brainliest
aleksklad [387]

Answer:

12 miles

Step-by-step explanation:

4*3 =12

7 0
3 years ago
Find the slope of the line. Write your answer in simplest form.
Savatey [412]

The slope of this line is: m = 3/4

3 0
3 years ago
Hi! I was wondering if you could help with this question please :)​
Makovka662 [10]

Answer:

R=\frac{QJ}{I^2t}

Step-by-step explanation:

So we have the equation:

Q=\frac{I^2Rt}{J}

And we want to solve for R.

First, let's multiply both sides by J to remove the fraction on the right. So:

(J)Q=(J)\frac{I^2Rt}{J}

Simplify the right:

JQ=I^2Rt

We can rewrite our equation as:

JQ=R(I^2t)

So, to isolate the R variable, divide both sides by I²t:

\frac{JQ}{I^2t}=\frac{R(I^2t)}{I^2t}

The right side cancels, so:

R=\frac{QJ}{I^2t}

And we are done!

7 0
3 years ago
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