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Anestetic [448]
3 years ago
8

What are the solutions to the equation (x – 2)(x + 5) = 0?

Mathematics
2 answers:
Bess [88]3 years ago
7 0

your Answer:

C: x = 2, -5

nataly862011 [7]3 years ago
4 0

A product of two (or more) factor can be zero if and only if at least one of the factors is zero.

In other words, you cannot multiply two non-zero real numbers, and have zero as a result.

So, if we want the product of these two factors to be zero, at least one of them has to be zero.

The first factor is zero if

x-2 = 0 \iff x=2

The second factor is zero if

x+5 = 0 \iff x=-5

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Arturiano [62]
Hey how are you doing today?
8 0
2 years ago
Read 2 more answers
Find the missing side length image below
barxatty [35]

Answer:

40

Step-by-step explanation:

Based on the Proportional Transversal Theorem, the three parallel lines hat intersects the two transversals, divides the transversal lines proportionally.

Therefore, we would have the following ratio:

28/35 = ?/50

Cross multiply

35*? = 50*28

35*? = 1,400

Divide both sides by 35

? = 1400/35

? = 40

4 0
3 years ago
HELPPPPPPP PLEASEEEEE !!!!
antiseptic1488 [7]

Answer:

4

Step-by-step explanation:

We know that all lengths in a rhombus are equal so we make the 2 equations equal to each other, then solve for x:

4x-6=5x-10

5x-4x=10-6

x=4

x=4

3 0
2 years ago
84.87 is 115% of what
elena-s [515]
Divide 84.87 by 1.15, and you get 73.8

Then you can check your work by multiplying 73.8 by 1.15 and you get 84.87
6 0
3 years ago
Evaluate ∫ xe2x dx. 1 2 3x A./xe +C 6 B.1/xe2x-1/ xe2x+C 22 C.1/xe2x-1/ e2x+C 24 1 2 1 4x D./x-/e +C 28
Dafna1 [17]
The answer is (1/2)xe^(2x) - (1/4)e^(2x) + C

Solution:
Since our given integrand is the product of the functions x and e^(2x), we can use the formula for integration by parts by choosing
     u = x
     dv/dx = e^(2x)

By differentiating u, we get
     du/dx= 1
By integrating dv/dx= e^(2x), we have
     v =∫e^(2x) dx = (1/2)e^(2x)

Then we substitute these values to the integration by parts formula:
     ∫ u(dv/dx) dx = uv −∫ v(du/dx) dx 
     ∫ x e^(2x) dx = (x) (1/2)e^(2x) - ∫ ((1/2) e^(2x)) (1) dx
                          = (1/2)xe^(2x) - (1/2)∫[e^(2x)] dx
                          = (1/2)xe^(2x) - (1/2) (1/2)e^(2x) + C
where c is the constant of integration.

Therefore, 
     ∫ x e^(2x) dx = (1/2)xe^(2x) - (1/4)e^(2x) + C
7 0
4 years ago
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