How do you find the value of x or y so that the line through the points has a given slope?

Answers

Answer 1

Answer: We can make a linear equation system to deduce it.

$\left \{ {{y_1=ax_1+b~(I)} \atop {y_2=ax_2+b~(II)}} \right. \n \n (II)-(I): y_2-y_1=a(x_2-x_1) \n \n \boxed {a= (\Delta x)/(\Delta y) }$

Answer 2

Answer: ok so if you had the points
(x1,y1) and (x2,y2)
slope=rise/run so
slope=(y2-y1)/(x2-x1)

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HELP ILL GIVE BRAINLY IF RIGHT !ROYGBIV is an
a.
Acronymic sentence
c.
Flashcard
b.
Acronymic abbreviation
d.
None of these

Please select the best answer from the choices provided

A
B
C
D

Answers

Answer:

Step-by-step explanation:

im pretty sure its c soo

Step-by-step explanation:

C because tjejkeksso akkaka

How many terms are present in the expression (2 + 5 + 8)?

Answers

3 terms are present in your equation: 2,5,8

Answer:

3.3.3.3.33.3.33.3.3.3

Step-by-step explanation:

Which expression is a difference of squares with a factor of 5x − 8?25x2 − 16
25x2 + 16
25x2 − 64
25x2 + 64

Answers

Answer:

option (3) is correct.

$25x^2-64$

Step-by-step explanation:

Given : factor 5x − 8

We have to choose an expression out of given option that is a difference of squares with a factor of 5x − 8

Using algebraic identity $a^2-b^2=(a+b)(a-b)$

a difference of squares is written mathematically as $a^2-b^2$

Since one factor is 5x − 8

using identity stated above other factor has be 5x+8

Thus $(5x+8)(5x-8)$

On simplify , we get

$(5x+8)(5x-8)=(5x)^2-8^2=25x^2-64$

Thus, option (3) is correct.

a2 - b2 = (a-b)(a+b)

(5x-8)(5x+8) = 25x2 - 64

answer: 25x2 - 64

The product of three and a squared number is twice the sum of the number and four

Answers

From the text of the task we can write the equation:

$3 x^(2) =2(x+4)\n \n 3x^2=2x+8\n \n 3x^2-2x-8=0\n \n \Delta=(-2)^2-4.3.(-8)=4+96=100\n \n x=(2 \pm√(100))/(2.3)=(2 \pm10)/(6)\n \n x_1=(2-10)/(6)=-(8)/(6)=-(4)/(3)\n \n x_2=(2+10)/(2.3)=(12)/(6)=2$

$3x^2=2(x+4)\n 3x^2=2x+8\n 3x^2-2x-8=0\n 3x^2-6x+4x-8=0\n 3x(x-2)+4(x-2)=0\n (3x+4)(x-2)=0\n x=-(4)/(3) \vee x=2$

Garland started the week with an odometer reading of 58,240 miles and a full tank. He ended the week with an odometer reading of 58,615, and he re-filled the tank with 15 gallons of gas. How many miles per gallon did he get to the nearest whole number? (Points : 2) 25
24
23
22

Answers

Answer:

A. 25.

Step-by-step explanation:

We have been given that Garland started the week with an odometer reading of 58,240 miles and a full tank. He ended the week with an odometer reading of 58,615, and he re-filled the tank with 15 gallons of gas.

To find the mileage for Garland's bike we will divide total distance traveled by total used fuel.

$\text{Mileage}=\frac{\text{Total distance}}{\text{Total amount of fuel}}$

$\text{Mileage}=\frac{\text{58,615 miles - 58,240 miles}}{\text{15 gallons}}$

$\text{Mileage}=\frac{\text{375 miles}}{\text{15 gallons}}$

$\text{Mileage}=\frac{\text{25 miles}}{\text{ gallon}}$

Therefore, Garland get a mileage of 25 miles per gallon and option A is the correct choice.

your answer is 25 miles per gallon!

Find the relative extrema of f(x) = 1+8x − 3x^2

Answers

f(x)=1+8x-3x²
1)we have to calculate the first derived
f´(x)=8-6x
2) we have to equalizate the first derived to "0", and find out the value of "x".
8-6x=0
-6x=-8
x=-8/-6=4/3
3)we have to calculate the second derived
f´´(X)=-6<0 ⇒ at x=4/3 we have a maximum.
4) we find the value of "y"
f(4/3)=1+8(4/3)-3(4/3)²=1+(32/3)-(48/9)=57/9

Therefore:
at (4/3 , 57/9) have a maximum

Other Questions

Simplify this expression (5)^ -5 (5)^ 7 please explain how you got the answer A.1/25B.25C.10D. 5^ -35

This week Jeff practiced playing the trombone for 1 hours onMonday and of an hour on Tuesday. On Wednesday, he practiced twice as long as he did on Tuesday. By the end of the day Thursday, he had practiced a total of 5 hours for the week. How long did he practice on Thursday? 3 3/4 hours 1 1/4 hours 2 hours 2 1/4 hours

Find the geometric mean x of each pair of numbers. 1.5 and 84

True or False: The value of r represents the reference angle when plotting a point in polar coordinates.