Find the line perpendicular to x=-7 through point (5,6)

Find the slope of the graph below and provide an explanation on how you found it.

ehidna [41]

Answer:

= -1/2

Step-by-step explanation:

Pick two points

(1,0) and (3,-1)

Then use the slope formula

m = (y2-y1)/(x2-x1)

= (-1 -0)/(3-1)

= -1 /2

7 0

3 years ago

Read 2 more answers

If KL = 3x, LM = 4, and KM = 5x, what is KL?

Anton [14]

KL is 3x

You sure you didn’t mistype it, the question gives you the value for KL

8 0

4 years ago

If you order these 3 from Least I will give 20 points and mark as brilliant 1.)6 3/5, 7 3/4,6.55,6,7.09 2.) 3 8/20,-3.3,3.41,3.2

natulia [17]

Answer:

1.) 6, 6.55, 6 3/5, 7.09, 7 3/4

2.) -3.3, -3 3/12, 3.27, 3 8/20, 3.41

3.) -0.4, 1/25, 4/50, 12/100, 0.8

Explanation:

1.) 6 3/5, 7 3/4, 6.55, 6, 7.09

6 3/5 -> 6.6

7 3/4 -> 7. 75

Therefore: 6, 6.55, 6 3/5, 7.09, 7 3/4

2.) 3 8/20, -3.3, 3.41, 3.27, -3 3/12

3 8/20 -> 3.04

-3 3/12 -> -3.25

Therefore: -3.3, -3 3/12, 3.27, 3 8/20, 3.41

3.) 4/50, 0.8, 12/100, -0.4, 1/25

4/50 -> 0.08

12/100 -> 0.12

1/25 -> 0.04

Therefore: -0.4, 1/25, 4/50, 12/100, 0.8

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

3 years ago

At a canning facility, a technician is testing a machine that is supposed to deliver 250 milliliters of product. The technician

Serhud [2]

Answer:

The value of the test statistic is $t = 1.97$

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 250$

The alternate hypotesis is:

$H_{1} \neq 250$

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the null hypothesis value, $\sigma$ is the standard deviation and n is the size of the sample.

In this problem:

$X = 251.6, \mu = 250, \sigma = 5.4, n = 44$

So

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$t = \frac{251.6 - 250}{\frac{5.4}{\sqrt{44}}}$

$t = 1.97$

The value of the test statistic is $t = 1.97$

4 0

4 years ago