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3 years ago

5

Find the solution set of each inequality below and then determine which inequalities have the same solution set as 1/3(-5x-3) &l

t;14

1 answer:

mario62 [17]3 years ago

5 0

1/3(-5x - 3) < 14

-5/3 x - 1 < 14

-5/3 x < 15

-5 x < 45

-x < 9

x > 9

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Kipish [7]

Answer:

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Step-by-step explanation:

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2 years ago

Find the standard form of the equation of the parabola with a focus at (0, -9) and a directrix y = 9.

ololo11 [35]

The Standard Form Is

x^2 = 0 (y - 9)

5 0

2 years ago

Read 2 more answers

———Help pleaseeeee———

Maslowich

Answer:

Slope of line is $\pi$

Step-by-step explanation:

Given that the company logo has four concentric circles.

To plot the line on the graph take point as ( Diameter, Circumference )

Circumference of a circle is given by C= $\pi D$

Where D is the diameter.

To draw a line, we need at least two-point.

Now,

Take D=1

Circumference of a circle will be C= $\pi D=\pi$

Required point is (1, $\pi$ )

Take D=0

Circumference of a circle will be C= $\pi D=0 \pi$

Required point is (0,0)

The slope of line is given by s= $\frac{Y2-Y1}{X2-X1}$

Hence,

s= $\frac{Y2-Y1}{X2-X1}$

s= $\frac{0-[tex]\pi$ }{0-1}[/tex]

s= $\frac{[tex]\pi$ }{1}[/tex]

s= $\pi$

Thus, Slope of line is $\pi$

7 0

3 years ago

Consider a line passing through the points A(–28, –13) and B(28, 15). Type the y-value for the point C(–24, y) to ensure that po

Nostrana [21]

Step-by-step explanation:

$\frac{y_2-y_1}{x_2-x_1}=\frac{15-(-13)}{28-(-28)}\\=\frac{28}{2(28)}\\\therefore\ m=\frac{1}{2}\\\frac{y-y_1}{xl-x_1}=m]\\\frac{y+13}{x+28}=\frac{1}{2}\\2y+26=x+28\\2y=x+2\\ y=\frac{1}{2}x+1$

In order to find y for point C on AB, substitute point C in line equation if AB.

$y=\frac{1}{2}(-24)+1\\\therefore y=-12+1=11\\\therefore C(-24, -11)$

7 0

3 years ago

A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and

Naya [18.7K]

<h2>The number of small buses used = 5</h2><h2>The number of big buses used = 4</h2>

Step-by-step explanation:

Let us assume the total number of small buses needed = x

The capacity of 1 small bus = 40

So, the capacity of x buses = 40(x) = 40 x

Let us assume the total number of big buses needed = y

The capacity of 1 big bus = 50

So, the capacity of y buses = 50(y) = 50 y

Also, the total students travelling = 400

So, the number of students traveling by (Small bus + Big bus) = 400

⇒ 40 x + 50 y = 400 ..... (1)

Also, the total number of drivers available = 9

⇒ x + y = 9 ..... (2)

Also, x ≤ 8, y ≤ 10

Now, solving both equations, we get:

40 x + 50 y = 400 ..... (1)

x + y = 9 ⇒ y = (9-x) put in (1)

40 x + 50 y = 400 ⇒ 40 x + 50 (9-x) = 400

or, 40 x + 450 - 50 x = 400

or, - 10 x =- 50

or, x = 5 ⇒ y = (9-x) = 9- 5 = 4

Hence the number of small buses used = 5

The number of big buses used = 4

8 0

3 years ago