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3 years ago

Find 2x2 − 2z4 + y2 − x2 + z4 if x = −4, y = 3, and z = 2 can you help me

1 answer:

8 0

Alright, so plugging it in, we get 2(-4)^2=2(2)^4+3^2-(-4)^2+2^4. Use PEMDAS with parenthesis and exponents to then get (2)(16)+9-16+16. Multiplying 1 and 16, we get 32+41-16+16=73

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2 years ago

14 x 1 = 14 exemplifies which property

NNADVOKAT [17]

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3 years ago

Read 2 more answers

How to solve (-1,8) and (8,-8)

Volgvan

Step-by-step explanation:

1. If you are trying to find a linear equation from those two points, use the equation y2-y1 over x2-x1. y2-y1 just means the second point's y coordinate minus the first point's y coordinate (same goes for x2-x1).

2. So if you were to plug the coordinates into the equation, it would be: -8-8 over 8-(-1).

3. Solve to get -16/9 because -8-8=-16 and 8-(-1)=9, so -16/9. -16/9 is the slope of the line in the y=mx +b equation.

4. It would be written like y=-16/9x +b

5. Now we need to find b which is the y-intercept. To find this pick one of the points (we'll just do (-1,8)), and plug in the x and y coordinates and solve for b.

8=-16/9(-1) +b
multiply -16/9 by -1 which is 16/9
subtract from both sides for it to be 8-16/9 on the left side which is 6 2/9, and that is b

6. The complete equation is now y=-16/9x + 6 2/9

3 0

3 years ago

Which expression is the result of subtracting (Z2-3i) from (Z1 + 2)? A B C D

Reika [66]

Answer:

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4 0

3 years ago

Read 2 more answers

How many solutions are there to this equation 3x(x-4)+5-x=2x-7

stich3 [128]

Answer:

2 solutions

Step-by-step explanation:

I like to use a graphing calculator to find solutions for equations like these. The two solutions are ...

x = 1
x = 4

To solve this algebraically, it is convenient to subtract 2x-7 from both sides of the equation:

3x(x -4) +5 -x -(2x -7) = 0

3x^2 -12x +5 -x -2x +7 = 0 . . . . . eliminate parentheses

3x^2 -15x +12 = 0 . . . . . . . . . . . . collect terms

3(x -1)(x -4) = 0 . . . . . . . . . . . . . . . factor

The values of x that make these factors zero are x=1 and x=4. These are the solutions to the equation. There are two solutions.

Alternate method

Once you get to the quadratic form, you can find the number of solutions without actually finding the solutions. The discriminant is ...

d = b^2 -4ac . . . . where a, b, c are the coefficients in the form ax^2+bx+c

d = (-15)^2 -4(3)(12) = 225 -144 = 81

This positive value means the equation has 2 real solutions.

4 0

3 years ago