All categories

2 years ago

Write an equation of the line passing through the point (4, -2) that is parallel to the line - 3x – 4y = 5.

2 answers:

7 0

I think the answer would be ———— y + 2 = - 3/4 ( x - 4 ) .

So, the line is parallel to - 3x - 4y = 5 and the line passes through the point (4, -2).

So, what I did was I pulled up a graphing site on my Chromebook called DESMOS, maybe you could use this in the future to help you out a little bit.

I then graphed the line of - 3x - 4y = 5 or you could simplify it to y = -3/4x + 5/4 .

The slope of the line is 3/4 . You can find this by taking the points (-15, 10) and (5, -5) and doing this...

m = y2 - y1 divided by x2 - x1 which would look like -5 - 10 divided by 5 - -15 which would simplify to -15 over 20 which would get you a slope of -3/4.

Then you take the formula y - y1 = m( x - x1) and substitute in your variables.

It should look something like this y + 2 = - 3/4 ( x - 4) .

This answer is parallel to -3x - 4y = 5.

I’m sorry, I’m not very good at explaining, but I hope this helps you out.

elixir [45]2 years ago

6 0

Answer:

Step-by-step explanation:

Let's solve for x.

−3x−4y=5

Step 1: Add 4y to both sides.

−3x−4y+4y=5+4y

−3x=4y+5

Step 2: Divide both sides by -3.

−3x

−3

4y+5

−3

−4

−5

Answer:

−4

−5

You might be interested in

As the average daily temperature in New York City decreases, the number of persons who are observed wearing sweaters in the work

ruslelena [56]

When the average daily temperature in New York City decreases, the number of persons who are observed wearing sweaters in the workplace increases, this implies a negative correlation.

Negative correlation simply means the relationship between two variables where as one variable increases, the other variable will reduce.

Therefore, in a situation where the average daily temperature in New York City decreases and the number of persons who are observed wearing sweaters in the workplace increases, this illustrates a negative correlation.

Read related link on:

brainly.com/question/25705812

8 0

1 year ago

There are a total of 15 dimes and quarters in a covered cup. The value of all 15 coins totals 2.85

QveST [7]

Answer:

There are 6 dimes and 9 quarters.

Step-by-step explanation:

Represent the number of dimes and the number of quarters by d and q.

Then d + q = 15, or q = 15 - d.

Also, ($0.25)q + ($0.10)d = $2.85, or 5q + 2d = 57.

Substituding 15 - d for q, we get 5(15 - d) + 2d = 57, or

75 - 5d + 2d = 57, or

75 - 3d = 57, or

3d = 18. Thus, there are 6 dimes. Since q = 15 - d, q = 15 - 6 = 9.

There are 6 dimes and 9 quarters.

6 0

2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

2 years ago

Solve the problem. The systolic blood pressures of the patients at a hospital are normally distributed with a mean of 136mm Hg a

shtirl [24]

we know that the 90 percentile in Z-score is ≈ 1.28 then,
=136 ± 1.28(13.8)
=(118.336 , 153.664)
hope it helps

6 0

2 years ago

A marketing researcher wants to find a 96% confidence interval for the mean amount those visitors spend per person per day while

ki77a [65]

Answer:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=12$ represent the population standard deviation

n represent the sample size

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=1-0.96 =0.04$ . And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got $z_{\alpha/2}=2.054$ , replacing into formula (b) we got:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

5 0

2 years ago