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Andrew [12]
3 years ago
5

What is the y intercept of y=7-2x

Mathematics
1 answer:
neonofarm [45]3 years ago
4 0

Answer: It would be 7. (0,7)

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What is the product of -2x^3+8 and -4x
slega [8]

Answer:

In mathematics, a product is a number or a quantity obtained by multiplying two or more numbers together. For example: 4 × 7 = 28 Here, the number 28 is called the product of 4 and 7.

Step-by-step explanation:

hope this helps you :)

8 0
3 years ago
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Find the value of x.
daser333 [38]

Answer:

Step-by-step explanation:

hello :

7x-6 = 3x+14

7x-3x =14+6

4x=20

x=20/4

<h2>x=5</h2>
3 0
3 years ago
The graph below shows the annual cost (green) and sales (red) of a company. What is the break-even point?
s344n2d4d5 [400]

The break-even point for the graph is at 11 units.

Step-by-step explanation:

Break-even point refers to the point on the graph where either of the parameters of the graph intercepts each other. The corresponding location of the position where the intersection occurs gives the break-even point.

In the graph annual cost is plotted in green on the Y axis, while the sales are plotted on x-axis in red.

When we observe the graph carefully, we find that two-line intercepts. When the point at which interception occurs is extended on the x-axis, the point is 11 units, which gives us the break-even units.

Hence the point is 11 units.

3 0
3 years ago
Dylan has a fish tank at home. He knows its height, h, is one inch less than twice the width, w, and the length, l, of the fish
garik1379 [7]

Answer:

A.  

The monomial expression 2w represents the height of the fish tank.

Step-by-step explanation:

7 0
3 years ago
Bernoulli differential equation... y'+xy=xy^2
snow_lady [41]
y'+xy=xy^2\implies y^{-2}y'+xy^{-1}=x

Let z=y^{-1}, so that z'=-y^{-2}y'. Then the ODE becomes linear in z with

-z'+xz=x\implies z'-xz=-x

Find an integrating factor:

\mu(x)=\exp\left(\displaystyle\int-x\,\mathrm dx\right)=e^{-x^2/2}

Multiply both sides of the ODE by \mu:

e^{-x^2/2}z'-xe^{-x^2/2}z=-xe^{-x^2/2}

The left side can be consolidated as a derivative:

\left(e^{-x^2/2}z\right)'=-xe^{-x^2/2}

Integrate both sides with respect to x to get

e^{-x^2/2}z=e^{x^2/2}+C

where the right side can be computed with a simple substitution. Then

z=1+Ce^{x^2/2}

Back-substitute to solve for y.

y^{-1}=1+Ce^{x^2/2}\implies y=\dfrac1{1+Ce^{x^2/2}}
3 0
3 years ago
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