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3 years ago

What are easy ways to study for the EOG and NCFE?

1 answer:

4 0

I am an organized person, so i make a checklist of all my notes and study them in order and that helps me a lot for the EOGs

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Steve and Ben have some money in their wallets. If Steve gives Ben $3, Ben will have twice as much as Steve. If Ben gives Steve

Zigmanuir [339]

Answer:

Money owned by Steve = $11

Money owned by Ben = $13

Step-by-step explanation:

Let x denotes the money owned by Steve and y denotes the money owned by Ben.

Then, Steve gives $3 to Ben

Money left with Steve = x-3 and Ben = y+3

Now, Ben will have twice as much as Steve.

⇒ y+3 = 2(x-3) .............(1)

If Ben gives Steve $7

Money left with Steve = x+7 and Ben = y-7

Then, the amount Ben has will be one-third that of Steve’s.

⇒ y-7 = $\frac{1}{3}$ (x+7) ..........(2)

Solving equation (1) and (2) by elimination method, we get

x = 11 and y = 13

⇒ Money owned by Steve = $11

Money owned by Ben = $13

5 0

3 years ago

Can someone please help me find the value of x

Nadya [2.5K]

Answer:

its approximately 55 cos you subtract 13 from 92

Step-by-step explanation:

8 0

3 years ago

Which equation represents the data in this function table?

777dan777 [17]

Answer:

Step-by-step explanation:

because its easy

6 0

3 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$

We use the initial condition $y(0)=-12$ to find the value of C.

$-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}$

Therefore,

$y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}$

4 0

3 years ago

Zane calculated the missing side length of one of these triangles using the Pythagorean Theorem. Which triangle is it?

topjm [15]

The pythagorean theorem states that for any RIGHT-ANGLED TRIANGLE, the relationship between the side lengths are A²+B²=C², where C is the hypotenuse :P

Which means that it can only be applied on right angled triangles~
So your answer would be B, a right-angled triangle.

Ask me if you are unsure about anything <3

6 0

3 years ago

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