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Sophie [7]
3 years ago
11

Which would not be a step in solving 5x + 15 + 2x = 24 + 4x?

Mathematics
2 answers:
Alona [7]3 years ago
8 0
Use distributive property.

Hope this helps!
motikmotik3 years ago
4 0
D use the distributive property 
your trying to find the variable 
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How do you simplify 2/(square root 2 + square root 2)
larisa [96]

Note that √2 + √2 = 2√2

2/(2√2) = (√2)/2 = 0.707 (simplified)

0.707 is your answer

~

7 0
2 years ago
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H=4(x+3y)+2 make x the subject
blagie [28]
H = 4(x + 3y) + 2
H = 4x + 12y + 2
H - 12y - 2 = 4x
(H - 12y - 2) / 4 = x or 1/4H - 3y - 1/2 = x

5 0
2 years ago
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PLEASE HELP WITH THIS I HAVE NO IDEA HOW TO SOLVE!!!!
VARVARA [1.3K]
Write the vertex form of the equation and find the necessary coefficient to make it work.
.. y = a*(x +3)^2 -2
.. = ax^2 +6ax +9a -2

You require the y-intercept to be 7. So, for x=0, you have
.. 9a -2 = 7
.. 9a = 9
.. a = 1

The equation you seek is
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4 0
2 years ago
What are a) the ratio of the perimeters and b) the ratio of the ratio of the areas of the larger figure to the smaller figure
vfiekz [6]
Assuming that the figures given are square such that the scale factor between them is equal to 28/8 which can be further simplified into 7/2. The ratio of the perimeter is also equal to this value, 7/2. However, the ratio of the areas is equal to the square of this value giving us an answer of 49/4. 
8 0
2 years ago
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Find the volume of the solid of revolution generated by revolving the region bounded by y = 2x^2, y = 0, and x = 2 about the x-a
Irina18 [472]

Answer:

128 \pi/5 units^3

Step-by-step explanation:

The volume of the solid revolution is expressed as;

V = \int\limits^2_0 {\pi y^2} \, dx

Given y = 2x²

y² = (2x²)²

y² = 4x⁴

Substitute into the formula

V = \int\limits^2_0 {4\pi x^4} \, dx\\V =4\pi \int\limits^2_0 { x^4} \, dx\\V = 4 \pi [\frac{x^5}{5} ]\\

Substituting the limits

V = 4 \pi ([\frac{2^5}{5}] - [\frac{0^5}{5}])\\V = 4 \pi ([\frac{32}{5}] - 0)\\V = 128 \pi/5 units^3

Hence the volume of the solid is 128 \pi/5 units^3

5 0
2 years ago
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