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aliina [53]
2 years ago
12

Please help me I will give you extra points and whoever the first one to answer I mark you with the brain thing. #5

Mathematics
1 answer:
bagirrra123 [75]2 years ago
6 0

Answer:

c

Step-by-step explanation:

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What is -12 as an integer
Serhud [2]
- 12. What do you mean by that question? -12 itself is an integer.
8 0
3 years ago
Substitution<br> 1.y=-5x-19<br> y=-6x-24<br><br> 2.y=-6x-22<br> y=-3x-10
omeli [17]

Answer:

1. y=5 and x=-5

2. y=2 and x=-4

Step-by-step explanation:

1.y=-5x-19

y=-6x-24

since both equations are equal to y, they are also equal to eachother

-5x-19=-6x-24

addition P.O.E.

x-19=-24

addition P.O.E.

x=-5

substitute x into one equation

y=-5*-5-19

y=6

2. y=-6x-22

y=-3x-10

since both equations are equal to y, they are also equal to eachother

-6x-22=-3x-10

addition P.O.E.

-22=3x-10

addition P.O.E.

-12=3x

division P.O.E.

-4=x

substitute x into one equation

y=-3*-4-10

y=2

5 0
2 years ago
What two numbers multiply to 72 and add to 14?
Ymorist [56]

Answer:

4 and -18

Step-by-step explanation:

6 0
2 years ago
Calculus question?
Ann [662]
Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)

Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2   Now differentiate that. It should be much easier.

Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x  I wonder if there's anything else you can do to this. If there is, I don't see it.

I suppose this is possible.
y' = 3/x^(1/2) + 6x

y' = \frac{3 + 6x^{3/2}}{x^{1/2}}

Frankly I like the first answer better, but you have a choice of both.
5 0
2 years ago
30 POINTS AVAILABLE
kherson [118]

Answer:

\large\boxed{(x-2)^2+(y-1)^2=34}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)

Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

Finally we have:

(x-2)^2+(y-1)^2=(\sqrt{34})^2

3 0
3 years ago
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