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krek1111 [17]
4 years ago
7

The slope of the line below is -5. Which of the following is the point slope form of the line (2,-7)

Mathematics
1 answer:
tatyana61 [14]4 years ago
7 0

Plugging in (2,-7) into

y=-5x+b, we get that b=3.

Therefore, the equation is y=-5x+3

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There are 9 numbers written, beginning with: 8, 5, 4, 9, 1, ... Finish the sequence.
alexdok [17]

Answer:

14    -50    -29    -541

Step-by-step explanation:

8-3=5-1=4+5=9-8=1+13=14-64=-50+21=-29-512=-541  

i got this by looking up sequence pattern finder in google and clicking on the second option then inserting the numbers you gave hope this helps

4 0
3 years ago
Find the area of the rectangle with the length 2x-4 and height of -3
denis-greek [22]

Answer:

-3(2x-4) or -6x+12

Step-by-step explanation:

Area is H*W

theres no value for x so that would be it, unless there is then all you do is plug that into the answer I put.

8 0
2 years ago
What are the zeros of the polynomial function f(x) = x3 − x2 − 12x?
kompoz [17]

Answer:

x = 0, x = 4, x = -3

Step-by-step explanation:

Factor out the x, which makes it x(x^2-x-12)

Then it will be x(x-4)(x+3)

x = 0

x - 4 = 0, x = 4

x + 3 = 0, x = -3

7 0
3 years ago
Hi guys, Can anyone help me with this tripple integral? Thank you:)
OleMash [197]

I don't usually do calculus on Brainly and I'm pretty rusty but this looked interesting.

We have to turn K into the limits of integration on our integrals.

Clearly 0 is the lower limit for all three of x, y and z.

Now we have to incorporate

x+y+z ≤ 1

Let's do the outer integral over x.  It can go the full range from 0 to 1 without violating the constraint.  So the upper limit on the outer integral is 1.

Next integral is over y.  y ≤ 1-x-z.   We haven't worried about z yet; we have to conservatively consider it zero here for the full range of y.  So the upper limit on the middle integration is 1-x, the maximum possible value of y given x.

Similarly the inner integral goes from z=0 to z=1-x-y

We've transformed our integral into the more tractable

\displaystyle \int_0^1 \int_0^{1-x} \int _0^{1-x-y} (x^2-z^2)dz \; dy \; dx

For the inner integral we get to treat x like a constant.

\displaystyle \int _0^{1-x-y} (x^2-z^2)dz = (x^2z - z^3/3)\bigg|_{z=0}^{z= 1-x-y}=x^2(1-x-y) - (1-x-y)^3/3

Let's expand that as a polynomial in y for the next integration,

= y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3

The middle integration is

\displaystyle \int_0^{1-x} ( y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3)dy

= y^4/12 + (x-1)y^3/3+ (2x+1)y^2/2- (2x^3+1)y/3 \bigg|_{y=0}^{y=1-x}

= (1-x)^4/12 + (x-1)(1-x)^3/3+ (2x+1)(1-x)^2/2- (2x^3+1)(1-x)/3

Expanding, that's

=\frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1)

so our outer integral is

\displaystyle \int_0^1 \frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1) dx

That one's easy enough that we can skip some steps; we'll integrate and plug in x=1 at the same time for our answer (the x=0 part doesn't contribute).

= (5/5 + 16/4 - 36/3 + 16/2 - 1)/12

=0

That's a surprise. You might want to check it.

Answer: 0

6 0
3 years ago
Give the values of a, b, and c needed to write the equation's general form. 2/3(x - 4)(x + 5) = 1
ale4655 [162]

The first step is to write this equation into general form. The general form of an equation is:

ax^2 + bx + c = 0

To make this equation to general form, you have to simplify the equation first.

2/3(x-4) (x+5) = 1

2/3 (x^2 + 5x – 4x – 20) = 1

2/3(x^2 + x -20) = 1

2/3x^2 + 2/3x – 40/3 = 1

2/3x^2 + 2/3x – 40/3 – 1 = 0

2/3x^2 +2/3x – 43/3 = 0

Therefore, a = 2/3 ; b = 2/3 ; c = -43/3   

4 0
3 years ago
Read 2 more answers
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