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2 years ago

What is 5(2 + x) = 5 ?

Mathematics

2 answers:

Alex787 [66]2 years ago

8 0

5(2+x)=5

or, 10+5x=5

or, 5x= -5

or, x= -1

Hope it helps

ser-zykov [4K]2 years ago

6 0

Answer:

−

Step-by-step explanation:

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2617 to the nearest thousand

Sliva [168]

Answer:

3000

Step-by-step explanation:

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2 years ago

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What percentage of the distance covered by an Olympic triathlon do athletes cycle if the swim for 3%, run for 20% and cycle the

Dafna1 [17]

100- (20+3)
=100-23
=77%

4 0

3 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

2 years ago

Domain and Range for the function f(x)=5IXI is

shutvik [7]

Answer:

The domain of the function f(x) is:

$\mathrm{Domain\:of\:}\:5\left|x\right|\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

The range of the function f(x) is:

$\mathrm{Range\:of\:}5\left|x\right|:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

Step-by-step explanation:

Given the function

$f\left(x\right)=5\left|x\right|$

Determining the domain:

We know that the domain of the function is the set of input or arguments for which the function is real and defined.

In other words,

Domain refers to all the possible sets of input values on the x-axis.

It is clear that the function has undefined points nor domain constraints.

Thus, the domain of the function f(x) is:

$\mathrm{Domain\:of\:}\:5\left|x\right|\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

Determining the range:

We also know that range is the set of values of the dependent variable for which a function is defined.

In other words,

Range refers to all the possible sets of output values on the y-axis.

We know that the range of an Absolute function is of the form

$c|ax+b|+k\:\mathrm{is}\:\:f\left(x\right)\ge \:k$

$k=0$

Thus, the range of the function f(x) is:

$\mathrm{Range\:of\:}5\left|x\right|:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

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2 years ago

What are the resulting coordinates of triangle A′B′C′ after reflecting triangle ABC across the origin if the vertices of triangl

Roman55 [17]

Answer:

Step-by-step explanation:

When reflecting across the origin, negetives become positives and positives become negetives.

Hope that helps!

6 0

2 years ago