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Damm [24]
4 years ago
13

-7 -5(-4c+4) <-7c+8-3

Mathematics
2 answers:
muminat4 years ago
4 0

Answer:

c  <  32/27

Step-by-step explanation:

−7−5(−4c+4)<−7c+8−3

Step 1: Simplify both sides of the inequality.

20c−27<−7c+5

Step 2: Add 7c to both sides.

20c−27+7c<−7c+5+7c

27c−27<5

Step 3: Add 27 to both sides.

27c−27+27<5+27

27c<32

Step 4: Divide both sides by 27.

27c/27  <  32/27

c <  32/27

WITCHER [35]4 years ago
4 0
The answer is c < 32/27
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Zach painted 92 square feet in the same amount of time that Blaise painted
Alika [10]

Answer:

The answer to your question is 252 in²

Step-by-step explanation:

Data

Zach = 92 ft²

Blaise = 13500 in²

Who painted more and how many more square inches? = ?

Process

1.- Convert the square feet to square inches

            1 ft² ----------------- 144 in²

          92 ft² ----------------   x

            x = (92 x 144)/1

            x = 13248 in²

2.- Subtract the lower value to the higher value

Area = 13500 in² - 13248 in²

Area = 252 in²

3.- Conclusion

Blaise painted 252 in² more than Zach.

4 0
3 years ago
What is the inequality for this verbal description?
Hitman42 [59]

Answer:

A) d \geq 5s + 9

3 0
2 years ago
Read 2 more answers
HELP PLEASE ASAP<br><br><br> Solve for x in the diagram below
Leni [432]

Answer:

29

Step-by-step explanation:

(x+12) + 100 + x = 180

           2x + 112 = 180

Subtract 112 from both sides

                    2x = 58

Divide 2 from both sides

                      x = 29

3 0
3 years ago
Cuanto es 69 dividido entre 87647​
Zigmanuir [339]
I think it’s 1,270.246376811594 :)
8 0
3 years ago
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
<span><span>|d+1|</span>=2⋅3</span> => <span><span>|d+1|</span>=6</span>First solution:
<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
8 0
3 years ago
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