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2 years ago

Which graph represents the solution set of the system of inequalities?

Mathematics

1 answer:

vivado [14]2 years ago

6 0

Solve first for the solution of the inequalities. This can be done by replacing first the inequalities sign with the equal sign.

x + y = 1
2y = x - 4

The values of x and y from the system of linear equation are 2 and -1. This means that the intersection of the lines should be at point (2, -1).

Substitute 3 to x and determine the value of y from the second inequality.
2y ≥ x - 4

Substituting,
2y ≥ 3 - 4, y ≥ -1/2

Hence, the solution to this item should be the fourth one.

You might be interested in

Simplify 8 over negative 4 divided by negative 3 over 9. 6 −6 12 −12

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Given

$\frac{8}{-4} / \frac{-3}{9}$

Required

Solve

$\frac{8}{-4} / \frac{-3}{9}$

We start by converting the division sign to multiplication

$\frac{8}{-4} / \frac{-3}{9} =\frac{8}{-4} * \frac{-9}{3}$

Then, we divide -9 by 3

$\frac{8}{-4} / \frac{-3}{9}=\frac{8}{-4} * -3$

Then, we divide 8 by -4

$\frac{8}{-4} / \frac{-3}{9}=-2 * -3$

Multiply -2 and -3 to give final result

$\frac{8}{-4} / \frac{-3}{9} = 6$

Hence, $\frac{8}{-4} / \frac{-3}{9}$ is equivalent to 6

4 0

2 years ago

If f(x)= 2x-7, what is the equation for f^-1(x)

Citrus2011 [14]

F^-1(x)= (x+7)/2
it's the inverse so adding is subtracting and multiplication is divison

4 0

2 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

<em>See attachment for result of 1-VarStats</em>

8 0

3 years ago

Are 27.4 ones the same as 274 tenths

nadezda [96]

Yes. You can find that out by multiplying .1 by 274. It equals 27.4.

8 0

3 years ago

Prove the theorem (AB )^T= B^T. A^T

Lisa [10]

Answer:

$(AB)^T = B^T.A^T (Proved)$

Step-by-step explanation:

Given (AB )^T= B^T. A^T;

To prove this expression, we need to apply multiplication law, power law and division law of indices respectively, as shown below.

$(AB)^T = B^T.A^T\\\\Start, from \ Right \ hand \ side\\\\B^T.A^T = \frac{B^T.A^T}{A^T}.\frac{B^T.A^T}{B^T} (multiply \ through) \\\\ = \frac{A^{2T}.B^{2T}}{A^T.B^T} \\\\=\frac{(AB)^{2T}}{(AB)^T} \ \ (factor \ out \ the power)\\\\= (AB)^{2T-T} \ (apply \ division \ law \ of \ indices; \ \frac{x^a}{x^b} = x^{a-b})\\\\= (AB)^T \ (Proved)$

3 0

2 years ago