Simplify the following expression 5^2-4/5+2

Meow help me pls :(((

pentagon [3]

Answer:

a) 3x-5=16

b) 30cm

Step-by-step explanation:

a) 5(x-1)=2(x+8)

5x-5=2x+16

5x-2x-5=16

3x-5=16

b) solving the equation gives x=7

put x=7 into first line, 5(7-1)=30cm

to check we can place it into second one as well

2(7+8)=30cm

5 0

3 years ago

Erin gets her exercise by running. The graph shows the distances she covers in a given amount of time. How many hours does it ta

soldi70 [24.7K]

The correct answer is c. 2 $\frac{1}{2}$ hours

Explanation:

The graphic shows the relationship between the distance Erin covers by running and the time this requires. Additionally, to know the specific time that takes for Erin to run a specific distance or vice versa it is necessary to observe at which point the two variables intersect in the line of the graph.

According to this, Erin runs 25 miles (y-axis) in 2.5 hours (x-axis), which you can know because in the line shown by the graph the distance 25 miles that is in the middle of 20 and 30 miles intersects with 2.5 hours or the middle between 2 and 3. Additionally, this number can be expressed as 2 and a half hours, 2.5 or even 2 $\frac{1}{2}$ hours because the fraction represents half. This makes option C the answer.

6 0

3 years ago

How to find the distance? I need help with part B

just olya [345]

Answer:

99.96

Step-by-step explanation:

Your height is correct.

Part B is asking for the length of the hypotenuse.

76.6^2 + 64.25^2 = x^2

x^2 = 9995.6225

x = 99.96 or 99.98 depending on if you round to 76.6 or not

8 0

3 years ago

Read 2 more answers

Sole the given inequality and graph it on the number line. -2x - 5 < 3

Anon25 [30]

Answer:

your answer would be X>-4

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