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11Alexandr11 [23.1K]
3 years ago
12

X/8 > = -1.5 i forgot how to do this, mind explaining?

Mathematics
2 answers:
Alex17521 [72]3 years ago
7 0

Answer:

x>= -12

Step-by-step explanation:

x/8>=-1.5

*8      *8

x>= -12

marta [7]3 years ago
7 0

Answer:

The answer to that would be x ≥ -12.

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Answer:

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Step-by-step explanation:

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Cal had $15. He spent $3 on pens and $8 on a book.
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He spent $11 dollars minus tax
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(ED. 21) Analytic Geometry Unit Test..... #3 Which point is a solution of x2 + y2 &gt; 49 and y ≤ –x2 – 4?
sdas [7]

Answer:

We want to find a solution of the system:

x^2 + y^2 > 49

y ≤ –x^2 – 4

Here we do not have any options, so let's try to find a general solution.

First, we can remember that the equation of a circle centered in the point (a, b) and of radius R is:

(x - a)^2 + (y - b)^2 = R^2

If we look at our first inequality, we can write it as:

x^2 + y^2 > 7^2

So the solutions of the first inequality are all the points that are outside (because the symbol used is >) of the circle of radius R = 7 centered in the origin.

From the other equation, we would get:

y ≤ –x^2 – 4

This is parabola, anything that is in the graph of the parabola or below will be a solution for this inequality.

Then the solutions of the system, are the ones that are in the region of solutions for both inequalities.

You can see the graph below, where both regions are graphed. The intersection of these regions is the region of the solutions for the system of inequalities:

by looking at the graph, we can see a lot of points that are solutions, like:

(0, -10)

(0, -15)

(2, -10)

etc.

8 0
3 years ago
Can someone plz help me with this!
asambeis [7]

Answer:

(arranged from top to bottom)

System #3, where x=6

System #1, where x=4

System #7, where x=3

System #5, where x=2

System #2, where x=1

Step-by-step explanation:

System #1: x=4

2x+y=10\\x-3y=-2

To solve, start by isolating your first equation for y.

2x+y=10\\y=-2x+10

Now, plug this value of y into your second equation.

x-3(-2x+10)=-2\\x+6x-30=-2\\7x=28\\x=4

System #2: x=1

x+2y=5\\2x+y=4

Isolate your second equation for y.

2x+y=4\\y=-2x+4

Plug this value of y into your first equation.

x+2(-2x+4)=5\\x+(-4x)+8=5\\x-4x+8=5\\-3x=-3\\x=1

System #3: x=6

5x+y=33\\x=18-4y

Isolate your first equation for y.

5x+y=33\\y=-5x+33

Plug this value of y into your second equation.

x=18-4(-5x+33)\\x=18+20x-132\\-19x=-114\\x=6

System #4: all real numbers (not included in your diagram)

y=13-2x\\8x+4y=52

Plug your value of y into your second equation.

8x+4(13-2x)=52\\8x+52-8x=52\\0=0

<em>all real numbers are solutions</em>

System #5: x=2

x+3y=5\\6x-y=11

Isolate your second equation for y.

6x-y=11\\-y=-6x+11\\y=6x-11

Plug in your value of y to your first equation.

x+3(6x-11)=5\\x+18x-33=5\\19x=38\\x=2

System #6: no solution (not included in your diagram)

2x+y=10\\-6x-3y=-2

Isolate your first equation for y.

2x+y=10\\y=-2x+10

Plug your value of y into your second equation.

-6x-3(-2x+10)=-2\\-6x+6x-30=-2\\-30=-2

<em>no solution</em>

System #7: x=3

y=10+x\\2x+3y=45

Plug your value of y into your second equation.

2x+3(10+x)=45\\2x+30+3x=45\\5x=15\\x=3

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Answer:

If you're looking for H it could be 6

Step-by-step explanation:

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