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3 years ago

PLEASE ANSWER ASAP!!!

Mathematics

1 answer:

qwelly [4]3 years ago

5 0

Answer:

D, E

Step-by-step explanation:

It’s not A because the equation can not be any pair of numbers.

It’s not B because that is stupid logic.

It’s not C because (3,4) and (10,8) don’t work

It’s not F because there is 1 solution

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a circle has a diameter of 10cm what is the best proximation of this area use 3.14 to approximate for pi

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Hello!

The answer would be $75.8^{cm2}$

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Jose needs to enclose a rectangular section of his yard. The perimeter of the section is 27 feet, and the area is 35 square foot

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3 years ago

What is the value of the expression when n=3 -2n(5+n-8-3n)

igor_vitrenko [27]

-2(3)(5+3-8-3(3))
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3 years ago

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Find the solution of the initial value problem dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

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3 years ago

FIRST CORRECT ANSWER GETS BRAINLIEST AND 50 POINTS.

AnnyKZ [126]

Answer:

A) $5^3*5^{-4}=5^{3-4}=5^{-1}=\frac{1}{5}$ (Yes. It has a value between zero and one.)

B) $\frac{3^5}{3^{-6}}=3^{5-\left(-6\right)}=3^{5+6}=3^{11}=177147$ (No. It does not have a value between zero and one.)

C) $\left(\frac{1}{4}\right)^3* \left(\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^{3+2}=\left(\frac{1}{4}\right)^5=\frac{1^5}{4^5}=\frac{1}{1024}$ . (Yes. It has a value between zero and one.)

D) $\frac{\left(-7\right)^5}{\left(-7\right)^7}=\left(-7\right)^{5-7}=\left(-7\right)^{-2}=\frac{1}{\left(-7\right)^2}=\frac{1}{49}$ (Yes. It has a value between zero and one.)

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3 years ago

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