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GenaCL600 [577]
3 years ago
12

NEED ANSWERS NOW

Mathematics
1 answer:
IrinaK [193]3 years ago
3 0

The answer is choice C

On the left side we have something in the form a*b+a*c+a*d

On the right side we have something in the form a*(b+c+d)

All together, we have a*b+a*c+a*d = a*(b+c+d), where the letters a,b,c,d are to be replaced with the proper values (0.5, 0.5, 0.3, and 0.2 in that exact order).

The idea of distribution is to multiply the outer value 0.5 by each of the values inside the parenthesis (0.5, 0.3, and 0.2) and add up the products.

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If f(x) = 4x3 - 2x and g(x) = x3 + 6x2 + 3x - 7, then find f(x)<br> + g(x)
leva [86]

Answer:

5x^3+6x^2+x-7

Step-by-step explanation:

f(x) = 4x^3 - 2x and g(x) = x^3 + 6x^2 + 3x - 7,

f(x) + g(x)= 4x^3 - 2x+ x^3 + 6x^2 + 3x - 7

               4x^3 + x^3 + 6x^2 + 3x- 2x - 7

Combine like terms

5x^3+6x^2+x-7

4 0
3 years ago
Y = -x - 6 2x - 3y = -2 what is the value for y in the solution?
PtichkaEL [24]

Answer: if I’m solving for x the answer would be x=4

Step-by-step explanation:Solve for the first variable in one of the equations, then substitute the result into the other equation.

8 0
3 years ago
How do I round 3.047 to the nearest hundred and show the math
goblinko [34]
The second place to the right after the decimal is the hundredth, so basically, to round, you cut off the rest of the decimal places afterwards. If the number after the hundredth is at or above 5, you should make the hundredth 1 higher, but if it is 4 or below, you will just leave it as is. In this case, 3.047 rounds to 3.05. There really is not any math work to show, but just the explanation.<span />
3 0
3 years ago
Jimmy is trying to factor the quadratic equation $ax^2 + bx + c = 0.$ He assumes that it will factor in the form \[ax^2 + bx + c
Shkiper50 [21]

If

ax^2+bx+c=(Ax+B)(Cx+D)

then

ax^2+bx+c=ACx^2+(AD+BC)x+BD

\implies a=AC

If a is known and Jimmy wants to find A, then he has to know the value of C.

6 0
3 years ago
Read 2 more answers
Please help this is very important
alexandr1967 [171]

Answer:

Here's what I get.

Step-by-step explanation:

1. Rotate 180° about origin

The formula for rotation of a point (x,y) by an angle θ about the origin is  

x' = xcosθ  -  ysinθ

y' = ycosθ + xsinθ

If θ = 180°, sinθ = 0 and cosθ = -1, and the formula becomes

x' =   -x

y' =  -y

The rule is then (x, y) ⟶ (-x, -y).

H: (-3, -5) ⟶ (3, 5)

J: (-5, -3) ⟶ (5, 3)

Q: (0, -1) ⟶ (0, 1)  

The vertices of H'J'Q' are (3, 5), (5, 3), and (0, 1).

2. Rotation 90° counterclockwise about origin

cos90°  = 0 and sin90° = 1

x' = xcos90°  -  ysin90° = -y

y' = ycos90° + xsin90° =  x

The rule is then (x, y) ⟶ (-y, x).

B: (4, 5) ⟶ (-5, 4)

L: (5, 0) ⟶ (0, 5)

S: (2, 2) ⟶ (-2, 2)  

The vertices of B'L'S' are (-5, 4), (0, 5), and (-2, 2).

3. Rotation 90° clockwise about origin

cos(-90°) = 0 and sin(-90°) = -1

x' = xcos(-90°)  -  ysin(-90°) =  y

y' = ycos(-90°) + xsin(-90°) =  -x

The rule is then (x, y) ⟶ (y, -x).

F: (1, -5) ⟶ (-5, -1)

H: (-2, -3) ⟶ (-3, 2)

U: (-4, -5) ⟶ (-5, 4)  

The vertices of F'H'U' are (-5, -1), (-3, -2), and (-5, 4).

4. Rotate 180° about origin

The rule is (x, y) ⟶ (-x, -y).

J: (1, -1) ⟶ (-1, 1)

V: (2, 0) ⟶ (-2, 0)

Y: (5, -3) ⟶ (-5, 3)

R: (4, -3) ⟶ (-4, 3)  

The vertices of J'V'Y'R' are (-1, 1), (-2, 0), (-5, 3), and (-4, 3).

The figures below show your shapes before and after the rotations.

7 0
3 years ago
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