What percent of total hours was spent jogging?

brilliants [131]

So you need to solve for the x and y values...I personally would use a graph. The is this website called desmos, it can help you with this kind of problem. You can see that the green point which is (10, 44) is on the line.. Lets check our work...

y = 4x + 4
44 = 4(10) + 4
44 = 40 + 4
44 = 44
This is your answer. I hope this helps love! :)

4 0

3 years ago

Factoring Trinomials: Factor t^2-4x+3

vova2212 [387]

3 + t^2 - 4 x there no further factor.

3 0

3 years ago

The figure below shows a tent with wires attached to help stabilize it. The length of each wire is 8 feet greater than the dista

Anestetic [448]

Answer:

The length of 2 wires are 13 feet and 12 feet [i don't know if the 3rd side is a wire, if so, the length is 5 feet]

Step-by-step explanation:

<em>The image of the tent is attached.</em>

<em />

As we can see, it creates a right triangle with "x+8" side being the hypotenuse of the triangle.

The pythagorean theorem tells us:

Hypotenuse^2 = Leg^2 + Leg^2

THe hypotenuse is "x+8", the two legs are "x" and "x+7" respectively. Let's solve for x:

$(x+8)^2=(x+7)^2+x^2$

We expand and solve for x:

$(x+8)^2=(x+7)^2+x^2\\x^2+16x+64=x^2+14x+49+x^2\\x^2+16x+64=2x^2+14x+49\\x^2-2x-15=0\\(x-5)(x+3)=0\\x=5,-3$

Length can't be negative, so x = 5.

Hence, the 3 sides of the triangle are:

x = 5

x + 8 = 13

x + 7 = 12

The length of 2 wires are 13 feet and 12 feet

8 0

3 years ago

Does anyone know the answer to these two questions? thank you.

tamaranim1 [39]

Problem 24

Part 1

$\displaystyle \int \csc(x)\left(\sin(x)+\cot(x)\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}\left(\sin(x)+\frac{\cos(x)}{\sin(x)}\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}*\sin(x)+\frac{1}{\sin(x)}*\frac{\cos(x)}{\sin(x)}dx\\\\\displaystyle \int 1+\frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle \int 1 dx+\int \frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle x+C_1+\int \frac{1}{u^2}du \ \text{ ... where } u = \sin(x)\\\\\displaystyle x+C_1+\int u^{-2}du\\\\$

Part 2

$\displaystyle x+C_1+\frac{1}{1+(-2)}u^{-2+1}+C_2\\\\\displaystyle x+C_1+\frac{1}{-1}u^{-1}+C_2\\\\\displaystyle x+C_1-u^{-1}+C_2\\\\\displaystyle x+C_1-\frac{1}{u}+C_2\\\\\displaystyle x+C_1-\frac{1}{\sin(x)}+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C_1+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C\\\\\displaystyle x-\csc(x)+C\\\\$

<h3>Answer: x - csc(x) + C</h3>

Don't forget about the plus C constant

==========================================================

Problem 26

Fortunately, there aren't as many steps for this problem.

$\displaystyle \int \frac{dy}{\csc(y)}\\\\\displaystyle \int \frac{1}{\csc(y)}dy\\\\\displaystyle \int \sin(y)dy\\\\\displaystyle -\cos(y)+C\\\\$

<h3>Answer: -cos(y) + C</h3>

6 0

3 years ago

Can someone find the inverse and explain it step to step please!!!!!!! (FOR QUESTION 3)

Klio2033 [76]

Answer:

f^-1(x) = 4(x+5)/3

Step-by-step explanation:

Let f(x) = y

So, y = (3/4)x - 5

y + 5 = (3/4)x

4(y+5)/3 = x

The steps are first to make a variable that holds the same value as the function, then from there you simplify and eventually make x the subject of the equation.

4 0

3 years ago