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1 year ago

In Exercises 1-16, solve the inequality. Graph the solution 1. 3y ≤ -9

Mathematics

1 answer:

murzikaleks [220]1 year ago

8 0

By solving the inequality we get $y\leq -3$

What is graphing the inequality?

The act of illustrating which area of the number line contains values that will "satisfy" the specified inequality is known as graphing the inequality. Take a look at the first inequality, "x > -5." The numbers that can be used to substitute x in our inequality to produce a true statement can be shown on a graph of our inequality.

$3y\leq -9$

Dividing above equation by 3 we get,

$\frac{3y}{3}\leq \frac{-9}{3}\\y\leq -3$

Graph of $y\leq -3$ is

Therefore by solving the inequality we get $y\leq -3$

To learn more about graphing the inequality from the given link

brainly.com/question/24372553

#SPJ1

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The value of X in the figure ABCD witch is a rhombus

NeX [460]

Answer:

x = 10°

Step-by-step explanation:

The diagonals of a rhombus divide the figure into congruent right triangles. That means the two marked acute angles are complementary.

<h3>solve for x</h3>

Complementary angles total 90°, so the relation we can use to find x is ...

(3x -13) +(8x -7) = 90

11x = 110 . . . . . . . . . add 20

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The value of x is 10°.

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2 years ago

What is the domain and range

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Answer:

The domain is all the x-values, and the range is all the y-values

Step-by-step explanation:

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3 years ago

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

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3 years ago

Read 2 more answers

Given the list of terms 1/3, 1, 5/3, 7/3,.... Find the 15th and −16th term.

qwelly [4]

Answer:

15th term =29/3

16th term = 31/3

Step-by-step explanation:

Given an arithmetic sequence with the first term a1 and the common difference d , the nth (or general) term is given by an=a1+(n−1)d .

First we find the 15th term

n=15

a1=1/3

d=1 - 1/3 = 2/3

Solution

1/3+(15-1)2/3

1/3+28/3

(1+28)/3

29/3

Lets find the 16th term

1/3+(16-1)2/3

1/3+30/3

(1+30)/3

31/3

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3 years ago

The basketball boosters are selling school t-shirts for 20 dollars and pants for 45 dollars. Last week they sold 100 items and c

lana66690 [7]

He sold 76 shirts and 24 pants.

Step-by-step explanation:

Given,

Cost of one t-shirt = $20

Cost of one pants = $45

Total items sold = 100

Total sales = 2600

Let,

x be the number of t-shirts

y be the number of pants

According to given statement;

x+y=100 Eqn 1

20x+45y=2600 Eqn 2

Multiplying Eqn 1 by 20

$20(x+y=100)\\20x+20y=2000\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(20x+45y)-(20x+20y)=2600-2000\\20x+45y-20x-20y=600\\25y=600$

Dividing both sides by 25

$\frac{25y}{25}=\frac{600}{25}\\y=24$

Putting in Eqn 1

$x+24=100\\x=100-24\\x=76$

He sold 76 shirts and 24 pants.

Keywords: linear equations, subtraction

Learn more about linear equations at:

#LearnwithBrainly

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3 years ago