A line of 3 passes through the point (5,13) what is the equation of the line

The length of a rectangle is (6x2 + 3x - 2) units, and its width is (x3 - 2x + 5) units. Part A: What is the area of the rectang

strojnjashka [21]

<span>Part A: Area = length * width = (6x^2 + 3x - 2) * (x^3 - 2x + 5) Multiply it out and simplify.

part B: </span><span>Take the first term 6x^2 and multiply each of the term x^3, -2x & 5. Then take 3x and multiply each of the term x^3, -2x & 5. Do the same with -2. Then add like terms and simplify.</span>

6 0

3 years ago

Write the standard form of the line that passes through (-1,-3) and (2,1).

kobusy [5.1K]

Answer:

Step-by-step explanation:

$\mathrm{Line\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Find\:the\:line\:}\\\mathbf{y=mx+b}\mathrm{\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Compute\:the\:slope\:}\left(-1,\:-3\right),\:\left(2,\:1\right):\quad \\m=\frac{4}{3}\\\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}\\\\\left(x_1,\:y_1\right)=\left(-1,\:-3\right),\:\left(x_2,\:y_2\right)\\\left\\$

$\\m=\frac{1-\left(-3\right)}{2-\left(-1\right)}\\\\Refine\\\\m=\frac{4}{3}\\\\\mathrm{Compute\:the\:}y\mathrm{\:intercept}:\quad b=-\frac{5}{3}\\\\y=\frac{4}{3}x-\frac{5}{3}\\$

Convert equation to standard form

$y - \frac{4}{3} x + \frac{5}{3} = 0\\\\Multiply \:through\: by\: 3;\\\\3\times (y - \frac{4}{3} x + \frac{5}{3} )=0\\\\3y -4x+5 = 0\\Arrange\:n\:form\:of \:;\\Ax +By = C\\\\-4x +3y =-5$

5 0

4 years ago

If a rectangle is inscribed in a circle, each diagonal also serves another function. What is this function?

Kipish [7]

Answer:

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle.

Step-by-step explanation:

A rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle. Hope this helps

7 0

3 years ago

Read 2 more answers

It snows 2 inches an hour for 5 hours in Pittsburgh, PA. How many inches fell in total over those 5 hours?

amm1812

Answer:

10 inches

Step-by-step explanation:

6 0

3 years ago

A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \

Yanka [14]

Answer:

a) $P(A)=0.875$

b) $\text{P(A or B)}=0.875$

c) $\text{P((not A) or B or (not C))}=0.625$

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

$P(A)=\frac{n(A)}{n(S)}$

$P(A)=\frac{7}{8}$

$P(A)=0.875$

b) P(A or B)

Using formula,

$\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}$

$\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}$

$\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}$

$\text{P(A or B)}=\frac{7}{8}$

$\text{P(A or B)}=0.875$

(c) P((not A) or B or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

$\text{P((not A) or B or (not C))}=\frac{5}{8}$

$\text{P((not A) or B or (not C))}=0.625$

4 0

3 years ago