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nikdorinn [45]
3 years ago
13

What’s this equation? 50 + 6x + 1 – 22x

Mathematics
1 answer:
Serga [27]3 years ago
6 0
In simplified form, the answer would be -16x+51
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1/5 ÷ 7 = ______________________________
joja [24]
0.02857 hope this helps
6 0
4 years ago
Read 2 more answers
How would u do this need help
Lunna [17]
Standard form: y = x^2 + 10x + 21
Axis of symmetry: x = -5
X intercepts: (-3,0) (-7,0)
Y intercepts: (0,21)
Vertex: (-5,-4)
4 0
4 years ago
Solve 2(1 – x) > 2x.
Fantom [35]

First you must distribute the 2 to the numbers inside the parentheses, which would be 1 and -x...

(2 * 1) + (2 * -x) > 2x

2 + (-2x) > 2x

2 - 2x >2x

Add 2x to both sides (what you do on one side you must do to the other). Since 2x is being subtracted, addition (the opposite of subtraction) will cancel it out (make it zero) from the left side and bring it over to the right side.

2 - 2x + 2x > 2x + 2x

2 + 0 > 4x

2 > 4x

Next divide 4 to both sides to finish isolating x. Since 4 is being multiplied by x, division (the opposite of multiplication) will cancel 4 out (in this case it will make 4 one) from the right side and bring it over to the left side.

2/4 > 4x / 4

1/2 > x

or

x < 1/2

Hope this helped!

~Just a girl in love with Shawn Mendes

8 0
4 years ago
Read 2 more answers
SUBTRACT: (6x3 - 7x2) - (- 7x4 + 7x3 +<br> 8x2)
dexar [7]

Answer:

-5

Step-by-step explanation:

Following the order of PEMDAS, you will multiply what is in the parentheses, then subtract what is in the parentheses, then finally subtract the whole thing.

(6×3 - 7×2) - (-7×4 + 7×3 + 8×2)

(18 - 14) - (-28 + 21 + 16)

(4) - (9)

-5

4 0
3 years ago
Find the coefficient of the squared term in the simplified form for the second derivative, f "(x) for f(x) = (x³ + 3x² + 3)(3x³
Yuki888 [10]
The coefficient for the equation is -24, because the formula applies like this

f(x) = (x^3 + 3x^2 + 3)(3x^3 - 6x^2 - 8x + 1)
     
      = +(x^3) (-8x) + (2x) (3x^3)

       Once we simplified through it all, we should be the answer

        f"(x) = -24x^2
5 0
3 years ago
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