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1 year ago

Graph the solution set of this inequality 3x-2y <_ 12

Mathematics

1 answer:

nadezda [96]1 year ago

8 0

The graph of the solution set for the inequality can be seen below.

<h3>How to graph the solution set?</h3>

Here we have the inequality:

3x - 2y < -12

If we isolate y, we get:

3x + 12 < 2y

(3x + 12)/2 < y

(3/2)x + 6 < y

Now, we just need to graph the line y = (3/2)x + 6 with a dashed line (because the points on the line are not solutions).

And then we need to shade the region above the line.

The graph of the solution set can be seen below.

If you want to learn more about inequalities:

brainly.com/question/18881247

#SPJ1

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Cerrena [4.2K]

Answer:

2x+h

Step-by-step explanation:

f(x+h) means that wherever there's an x in the function f(x) we plug in a x+h

we have

(x+h)²+9

Let's start with the numerator

(x+h)²+9-(x²+9)

Distribute the exponent

x²+2xh+h²+9-x²-9

Which gives us

2xh+h²

Divide the whole thing by h to give us

2x+h

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2 years ago

The base length of a triangle is 8 feet and the height is 4 feet. What is the area of the triangle?

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3 years ago

Read 2 more answers

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ASHA 777 [7]

Answer:

As they triangles are congruent, we can say that, angle A = angle O

So, angle A = 105+25 =130

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2 years ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

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2 years ago