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2 years ago

A line has slope Four-thirds. Through which two points could this line pass?

Mathematics

1 answer:

lisabon 2012 [21]2 years ago

6 0

Answer:

Two points this line can pass through (1, 4/3) and (2, 8/3)

Step-by-step explanation:

The equation of the line is written in the slope-intercept form,

y = mx + b

where m represents the slope

b represents the y-intercept

where the format of a point is (x, y)

Given:

slope = 4 / 3
equation: y = 4/3x

we can plug in a number for x to find y

when x = 0,

y = 4/3(0)

y = 0

point: (0,0)

when x = 1,

y = 4/3(1)

y = 4/3

point: (1, 4/3)

when x = 2,

y = 4/3(2)

y = 8/3

point: (2, 8/3)

Learn more about slope and y-intercept here: brainly.com/question/12763756

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Write the product as a trinomial.(k – 1)(k – 4)

SashulF [63]

(k - 1)(k - 4)

Use FOIL to solve this.

First
k * k
Outer
k * -4
Inner
-1 * k
Last
-1 * -4

k^2 - 4k - 1k + 4

k^2 - 5k + 4 is the trinomial.

5 0

3 years ago

In July, the average temperature in one US city was 29 degrees by December, the average temperature had fallen by 29 degrees. Ex

VikaD [51]

Step-by-step explanation:

In July, it was positive 29 degrees, but in December it was negative 29 degrees. When you have 29-29 you get 0. Hope that helped

4 0

3 years ago

Read 2 more answers

Given: is an angle bisector of ∠JMK Prove: m

astraxan [27]

Answer:

An angle bisector is a line passing through the vertex of the angle that cuts the angle into two equal smaller angles.

Given: MN is angle bisector,

then

$\angle JMN \cong \angle NMK$ ....... [1]

Congruent angles are two or more angles that have the same measure.

then;

by definition of congruent angles

[1]⇒ $m\angle JMN = m\angle NMK$ ......[2]

By the Angle addition postulates states that if M is in the interior of ∠JMK then,

$m\angle JMN+m\angle NMK =m\angle JMK$ ......[3]

Now, by substitution property ; substitute the equation [2] in [3] we get;

$m\angle JMN+m\angle JMN =m\angle JMK$ ......[4]

Like terms terms whose variables are the same

Combine like terms in equation [4] we get

$2 \cdot m\angle JMN=m\angle JMK$ ......[5]

Division property of equality states that you divide the same number to both sides of an equation.

Divide by 2 to both sides in equation [5] , we get

$m\angle JMN= \frac{1}{2} m\angle JMK$

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3 years ago

An open box with a square base is to have a volume of 18 ft^3.

kotegsom [21]

Answer:

2.29 ft of side length and 1.14 height

Step-by-step explanation:

a) Volume V = x2h, where x is side of square base and h is hite.

Then surface area S = x2 + 4xh because box is open.

b) From V = x2h = 6 we have h = 6/x2.

Substitude in formula for surface area: S = x2 + 4x·6/x2, S = x2 + 24/x.

We get S as function of one variable x. To get minimum we have to find derivative S' = 2x - 24/x2 = 0, from here 2x3 - 24 = 0, x3 = 12, x = (12)1/3 ≅ 2.29 ft.

Then h = 6/(12)2/3 = (12)1/3/2 ≅ 1.14 ft.

To prove that we have minimum let get second derivative: S'' = 2 + 48/x3, S''(121/3) = 2 + 48/12 = 6 > 0.

And because by second derivative test we have minimum: Smin = (12)2/3 + 4(12)1/3(12)1/3/2 = 3(12)2/3 ≅ 15.72 ft2

5 0

2 years ago

9(2w−y)=21w−9y Solve for y

valentinak56 [21]

9(2w−y)=21w−9y

First you multiply the numbers in the parenthesis by 9

$18w-9y=21w-9y$

subtract. 21w-18w=3w

$-9y=3w-9y$

add. -9y+9y=0

$0=3w$

divide.

$0=w$

Final answer is 0.

5 0

3 years ago