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Elza [17]
3 years ago
6

Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answ

er in the box provided or use the upload option to submit your solution (-8,0) and (1,5)
Mathematics
1 answer:
dalvyx [7]3 years ago
7 0

The standard form is given as Ax+By =C

we are given points (-8,0) and (1,5)

let us first find the slope here

slope = \frac{y2-y1}{x2-x1}

slope = \frac{5-0}{1-(-8)}

slope=5/9

now using the equation (y-y1)=m(x-x1)

(y-0)=5/9(x+8)

9y=5x+40

9y-5x=40 is the standard form of the line that passes through given points

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Consider the equation y = 3x - 1. Identify the slope and y-intercept and
tatyana61 [14]

Answer: slope=3

           y intercept :(0, -1)

Step-by-step explanation:Use the slope-intercept form to find the slope and y-intercept

4 0
3 years ago
For which pair of numbers is 7 the greatest common factor (GCF)?
Elanso [62]

Answer:

B cuz both can be divided by 7

5 0
2 years ago
On a number line, what is the distance between -29 and 100? A) 29 B) 71 C) 81 D) 129
shusha [124]

-29 is 29 bars away from 0. 100 is 100 bars away from 0.

29 + 100 = 129

So the distance between -29 and 100 is 129. The answer is D.

6 0
2 years ago
A conical water tank with vertex down has a radius of 13 feet at the top and is 21 feet high. If water flows into the tank at a
VLD [36.1K]

Answer:

\frac{dh}{dt}\approx0.08622\text{ ft/min}

Step-by-step explanation:

We know that the conical water tank has a radius of 13 feet and is 21 feet high.

We also know that water is flowing into the tank at a rate of 30ft³/min. In other words, our derivative of the volume with respect to time t is:

\frac{dV}{dt}=\frac{30\text{ ft}^3}{\text{min}}

We want to find how fast the depth of the water is increasing when the water is 17 feet deep. So, we want to find dh/dt.

First, remember that the volume for a cone is given by the formula:

V=\frac{1}{3}\pi r^2h

We want to find dh/dt. So, let's take the derivative of both sides with respect to the time t. However, first, let's put the equation in terms of h.

We can see that we have two similar triangles. So, we can write the following proportion:

\frac{r}{h}=\frac{13}{21}

Multiply both sides by h:

r=\frac{13}{21}h

So, let's substitute this in r:

V=\frac{1}{3}\pi (\frac{13}{21}h)^2h

Square:

V=\frac{1}{3}\pi (\frac{169}{441}h^2)h

Simplify:

V=\frac{169}{1323}\pi h^3

Now, let's take the derivative of both sides with respect to t:

\frac{d}{dt}[V]=\frac{d}{dt}[\frac{169}{1323}\pi h^3}]

Simplify:

\frac{dV}{dt}=\frac{169}{1323}\pi \frac{d}{dt}[h^3}]

Differentiate implicitly. This yields:

\frac{dV}{dt}=\frac{169}{1323}\pi (3h^2)\frac{dh}{dt}

We want to find dh/dt when the water is 17 feet deep. So, let's substitute 17 for h. Also, let's substitute 30 for dV/dt. This yields:

30=\frac{169}{1323}\pi (3(17)^2)\frac{dh}{dt}

Evaluate:

30=\frac{146523}{1323}\pi( \frac{dh}{dt})

Multiply both sides by 1323:

39690=146523\pi\frac{dh}{dt}

Solve for dh/dt:

\frac{dh}{dt}=\frac{39690}{146523}\pi

Use a calculator. So:

\frac{dh}{dt}\approx0.08622\text{ ft/min}

The water is rising at a rate of approximately 0.086 feet per minute.

And we're done!

Edit: Forgot the picture :)

3 0
3 years ago
Critical Thinking: Data Transformation In this problem, we explore the effect on the standard deviation of multiplying each data
suter [353]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data : 5, 9, 10, 11, 15

s = √[(ΣX - m)² / (n - 1)]

n = number of sample = 5

m = mean

m = ΣX/n

m = (5 + 9 + 10+ 11 + 15) / 5

m = 50/5 = 10

s =√((5-10)^2 + (9-10)^2 + (10-10)^2 + (11-10)^2 + (15-10)^2) / 5-1

= 3.601

(B) Multiply each data value by 5 to obtain the new data set 25, 45, 50, 55, 75. Compute s.

m = ΣX/n

m = (25 + 45 + 50 + 55 + 75) / 5

m = 250/5 = 50

s =√((25-50)^2 + (45-50)^2 + (50-50)^2 + (55-50)^2 + (75-50)^2) / 5-1

= 18. 028

(c) Compare the results of parts (a) and (b). In general, how does the standard deviation change if each data value is multiplied by a constant c?

The standard deviation changes by almost c times multiplied by the initial value.

3.601 * 5 = 18.005 which is almost equivalent to 18.028

(d) You recorded the weekly distances you bicycled in miles and computed the standard deviation to be s 3.1 miles. Your friend wants to know the standard deviation in kilometers. Do you need to redo all the calcula- tions?

No

Given 1 mile 1.6 kilometers, what is the standard deviation in kilometers?

If 1 mile = 1.6km

s = 3.1 miles

s in kilometers = (1.6 * 3.1) = 4.96km

6 0
2 years ago
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