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2 years ago

Differnece between 2x+4 and 3x+4 and 4x+4

Mathematics

1 answer:

prohojiy [21]2 years ago

6 0

Answer:

the difference is the slope

Step-by-step explanation:

y = mx+b

m is the slope

b is the y-intercept

2x+4

3x+4

4x+4

the m constant changes meaning the slope is the difference

You might be interested in

POINTS X, Y and Z are COLLinear. You are given

Anon25 [30]

Answer:

129

Step-by-step explanation:

Collinear means that the points X, Y, and Z are on the same line. The question asks for one possible value of YZ, so for this, let's assume point X is between Point Y and Point Z.

This means...

XY+XZ=YZ

54+75=YZ

YZ=129

5 0

2 years ago

Round this number to nearest 10,000. 4,278,003

s2008m [1.1K]

Hi there!

To Round of the number given, first you have to find the ten thousands place on the given number. Started from ones, tens, hundreds, and moving on, the number in the ten thousands place would be 7 in your case.

Now, looking to the number on the right of 7 which is 8, it is bigger than 5, and therefore, we round up 7 and let 8 go to zero.

Applying this, your answer would be 4,280,003

Hope this helped!

4 0

2 years ago

Read 2 more answers

Need help ASAP!! Write an informal proof to show triangles ABC and DEF are similar.

yaroslaw [1]

Answer:

Δ ABC and Δ DEF are similar because their corresponding sides are proportional

Step-by-step explanation:

Two triangles are similar if their corresponding sides are proportional which means the corresponding sides have equal ratios

In the two triangles ABC and DEF

∵ AB = 4 units

∵ DE = 2 units

∴ $\frac{AB}{DE}=\frac{4}{2}=2$

∵ BC = 6 units

∵ EF = 3 units

∴ $\frac{BC}{EF}=\frac{6}{3}=2$

∵ CA = 2 units

∵ FD = 1 units

∴ $\frac{CA}{FD}=\frac{2}{1}=2$

∴ $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=2$

∵ All the ratios of the corresponding sides are equal

∴ The corresponding sides of the two triangles are proportional

∴ Δ ABC is similar to Δ DEF

6 0

2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

Substitution method need help for home work

Marizza181 [45]

-3y = x...so we sub in -3y for x in the other equation

-x + 7y = 70
-(-3y) + 7y = 70
3y + 7y = 70
10y = 70
y = 70/10
y = 7

so we sub in 7 for y in either of the original equations to find x
-3y = x
-3(7) = x
-21 = x

so ur solution is (-21,7)

6 0

2 years ago