S=Ut + 1\2 at^2

A rental moving truck can haul a maximum of 7,500 pounds. Dejuan has boxes that weigh 40 pounds and boxes that weigh 75 pounds.

Lelechka [254]

Answer:

One hundred and fifty 40 pounds boxes and twenty 75 pounds boxes

Step-by-step explaination:

The maximum amount he can haul in one load is 7500 pounds

150 × 40 pounds boxes = 6000 pounds

20 × 75 pounds boxes = 1500 pounds

6000 pounds + 1500 pounds = 7500 pounds

8 0

3 years ago

Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are neither parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are parallel

║

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are Neither parallel nor perpendicular

Neither

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are perpendicular

⊥

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are parallel

║

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are neither parallel nor perpendicular

Neither

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

Neither

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is Neither

Learn more here:

brainly.com/question/16732089

6 0

3 years ago

One of the zeros of the polynomial function is 5.

maks197457 [2]

The last term is -100 so its either a or d.
Try a:-

(x^2 - 25)(x-2)(x - 2)

= x^2 - 25 (x^2 - 4x + 4)

= x^2 + 4x^3 + 4x^2 - 25x^2 + 100x - 100
= x^4 + 4x^3 - 21x^2 + 100x - 100

Its a.

7 0

3 years ago

Read 2 more answers

Make a table of values for the equation. y = 3x + 2

Rzqust [24]

Tables are easy to make when u have the equation. Just simply sub any number in for x and solve for y

y = 3x + 2...lets let x = 0
y = 3(0) + 2
y = 2.....so when x = 0, y = 2....(0,2)

y = 3x + 2....lets let x = 1
y = 3(1) + 2
y = 3 + 2
y = 5....so when x = 1, y = 5....(1,5)

y = 3x + 2.....let x = 2
y = 3(2) + 2
y = 6 + 2
y = 8....so when x = 2, y = 8...(2,8)

u can find more point if u need to...

now make a table..

x y
0 2
1 5
2 8
and like I said, if u need more numbers for ur table, just do like I did above :)

7 0

3 years ago

Read 2 more answers

Which of the following is the least? A. 0.105 B. 0.501 C. 0.015 D. 0.15

Naddika [18.5K]

C is the least that is beause it is .015 which is smaller than 0.15

8 0

2 years ago