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3 years ago

Graph 3(8-4x)<6(x-5)

Mathematics

1 answer:

Zinaida [17]3 years ago

4 0

First simplify this inequality:

$3(8-4x)$

Now you can graph this inequality. Draw the vertical dashed line x=3 and shade the region to the right from this line. This is exactly the region that represents the solution of inequality (see attached diagram for details).

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Rose bought 5 pencils for $0.69 each, 3 notebooks for $ 2.75 each, and a pocket dictionary for

Marysya12 [62]

Answer: $2.81

Step-by-step explanation:

Pencils: 5 * 0.69 = 3.45

Notebooks: 3 * 2.75 = 8.25
pocket dictionary: 5.49

3.45 + 8.25 + 5.49 = 17.19

20 - 17.19 = 2.81

Therefore, she will get $2.81 change back.

3 0

2 years ago

Read 2 more answers

Which value, when placed in the box, would result in a system of equations with infinitly many solutions? Y=2x - 5 2y-4x=

Nezavi [6.7K]

Given:

The system of equations is

$y=2x-5$

$2y-4x=?$

To find:

The missing value for which the given system of equations have infinitely many solutions.

Solution:

Let the missing value be k.

We have,

$y=2x-5$

$2y-4x=k$

Taking all the terms on the left side, the given equations can be rewritten as

$-2x+y+5=0$

$-4x+2y-k=0$

The system of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ have infinitely many solutions if

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

We have,

$a_1=-2,b_1=1,c_1=5$

$a_2=-4,b_2=2,c_2=-k$

Now,

$\dfrac{-2}{-4}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{5}{-k}$

On cross multiplication, we get

$-k=10$

$k=-10$

Therefore, the missing value is -10.

6 0

3 years ago

Which equation does the graph of the systems of equations solve?

mihalych1998 [28]

Answer:

$\text{B. }\frac{1}{3}x-3=-x+1$

Step-by-step explanation:

Let's start by taking a look at the blue line. The slope of any line that passes through two points is equal to the change in y-values over the change in x-values. We can see that the line passes through points (0, 1) and (1, 0). Assign these points to $(x_1,y_1)$ and $(x_2, y_2)$ (doesn't matter which you assign) and use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Let:

$(x_1,y_1)\implies (0,1)\\(x_2,y_2)\implies (1, 0)$

The slope is equal to:

$m=\frac{0-1}{1-0}=\frac{-1}{1}=-1$

Therefore, the slope of this line is -1. In slope-intercept form $y=mx+b$ , $m$ represents slope, so one of the lines must have a term with $-x$ in it, which eliminates answer choices A and D.

For the second line, do the same thing. The red line clearly passes through (0, -3) and (3, -2). Therefore, let:

$(x_1,y_1)\implies (0,-3)\\(x_2,y_2)\implies (3, -2)$

Using the slope formula:

$m=\frac{-2-(-3)}{3-0}=\frac{1}{3}$

Thus, the slope of the line is 1/3 and the other line must have a term with $\frac{1}{3}x$ in it, eliminating answer choice C and leaving the answer $\boxed{\text{B. }\frac{1}{3}x-3=-x+1}$