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2 years ago

Find the slope of a line that passes through (7,-6) and (-5,-8)

Mathematics

1 answer:

AfilCa [17]2 years ago

3 0

Change in x over change in y

(y2-y1)/(x2-x1)
(-8--6)/(-5-7)
-2/-12
slope = 1/6

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suter [353]

Answer:

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1 year ago

How would you write a phrase that represents the expression 7(n - 12)?

NemiM [27]

7n-84 would be your correct answer, 7 times n is 7n. 12 times 7 is 84 so therefore your answer would be 7n-84

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2 years ago

Help me please?? solve using proportion

gulaghasi [49]

9514 1404 393

Answer:

2 1/2 cups

Step-by-step explanation:

A proportion says the ratio of one thing to another is the same for different amounts. Here, we want cups, so we can use the proportion ...

cups/people = 2/12 = x/15 . . . . . where x is the number we want to find

Multiplying by 15 gives ...

x = 15(2/12) = 5/2 = 2 1/2

You will need 2 1/2 cups of zucchini to feed 15 people with the same recipe.

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2 years ago

Find a rational number and an irrational number that are between 5.2 and 5.5. Include the decimal approximation of the irrationa

mote1985 [20]

Answer:

5.3

Step-by-step explanation:

3 0

1 year ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

2 years ago