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Levart [38]
2 years ago
12

For the following system, if you isolated x in the second equation to use the subsition method, what expression would you subsit

ute into the first equation?
2x + y = 8
-x - 3y = -12

A. 3y + 12
B. -3y + 12
C. 3y - 12
D. -3y - 12​
Mathematics
1 answer:
kkurt [141]2 years ago
7 0
A.3y+12

i think so,or maybe D or B
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Mai gave half of her brownies, and then half a brownie more, to Kiran. Then she gave half of what was left,
givi [52]

Answer:

x = 7

Step-by-step explanation:

all we know is that she had 1 brownie at the end, in the end she gave half her brownie to tyler, so add one who will get 1 1/2, before that she split her brownies between her friends, so multiply 1 1/2 by 2 and you will get 3, before that she gave 1/2 a brownie to Kiran, so we would add 1/2 to 3 and get 3 1/2, before that she first split it in half for a her friends, so at last,  multiply 3 1/2 by 2 and you will geet 7 as your final answer.

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Find all real square roots of 16.
Cerrena [4.2K]

4

that is the only one

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2 years ago
Help please this is so hard I don't understand it​
Zina [86]

Answer:

<ABC = 15

< BAC = 150

Step-by-step explanation:

<C = <B since this is an isosceles triangle.  We know this because AC = AB

<C = 15 so <ABC = 15

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A + B+ C = 180

A + 15+15 = 180

A +30 =180

A = 180-30

A = 150

< BAC = 150

5 0
3 years ago
Solve the equation for x 1/3(x-15) =95<br><br> A. X= 105<br> B. X= 215 <br> C. X= 300<br> D. X= 310
Korolek [52]
The answer is c x=300
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2 years ago
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Please help! I cannot figure this out, I have no idea what to do.
joja [24]

3. First you should identify which line corresponds to which equation.

x + 2y = -8   ⇒   y = -x/2 - 4

is the line with slope -1/2 and intercept (0, -4), so the first inequality refers to the lower line in the graph.

For the inequality itself, the solution set that satisfies it is either the region above or below it. To decide which, pick any point in the plane and plug its coordinates into the inequality. The origin is a natural choice, since it's not on the line and working with 0s is easy.

In the first inequality, we have

x = 0 and y = 0   ⇒   0 + 2•0 = 0 < -8

which is not true. So (0, 0) is not in the solution set, and it stands to reason that any point in the same region will not be a solution. In other words, the region above the line x + 2y = -8 does not satisfy the inequality, while the one below it does.

In the second inequality,

x = 0 and y = 0   ⇒   2•0 + 0 = 0 ≥ -1

which is true. This means the region containing (0, 0) above the upper line solves the second inequality.

The solution to the system of inequalities is then the intersection of these two regions. (See attached; I've labeled the solution to the first inequality in blue, the solution to the second one in red, and the solution to both in green.)

All this work is kind of overkill for this particular question, though. All you really need to do is check if the given point satisfies the inequalities:

• (-5, 5) :

x = -5 and y = 5   ⇒   -5 + 2•5 = 5 < -8   ⇒   NO

• (2, -5) :

x = 2 and y = -5   ⇒   -2 + 2•5 = 8 < -8   ⇒   NO

• (-4, -6) :

x = -4 and y = -6   ⇒   -4 + 2•(-6) = -16 < -8   ⇒   MAYBE

x = -4 and y = -6   ⇒   2•(-4) + (-6) = -14 < -8   ⇒   YES

• (5, -7) :

x = 5 and y = -7   ⇒   5 + 2•(-7) = -9 < -8   ⇒   MAYBE

x = 5 and y = -7   ⇒   2•5 + (-7) = 3 < -8   ⇒   NO

4. x is the number of candy boxes with chocolates and y is the number of boxes without chocolates. Each of the x boxes cost $27.50, so if you have x boxes, their total cost is $27.5x. Each of the y boxes cost $25.00, so y boxes cost $25y. The sponsor doesn't want to order more than 100 boxes total, so

x + y < 100

but wants to raise at least $2000, so

27.5x + 25y ≤ 2000

Now do the same thing as before; plug in the listed x- and y-coordinates and pick the point that satisfies both inequalities. You will find that (50, 30) is the correct choice.

5. Consult the "overkill" part of problem 3. The line y = -3x + 6 has a negative slope, so it's the downward sloping one. Check if the origin satisfies the inequality:

x = 0 and y = 0   ⇒   0 ≤ -3•0 + 6   ⇒   0 ≤ 6   ⇒   YES

This means Regions II and III solve the first inequality.

Do the same with the other inequality:

x = 0 and y = 0   ⇒   0 ≥ 1/2•0 + 1   ⇒   0 ≥ 1   ⇒   NO

This tells us that Regions I and II solve the second inequality.

The intersection of these regions is of course Region II.

6. One of the plotted lines is apparently y = x + 4, which has a positive slope, so this must be the upper line. The shaded region below it corresponds to the solution of either y < x + 4 or y ≤ x + 4 and we eliminate B.

The other line has negative slope, which eliminates A and D since

3x - 4y = 20   ⇒   y = 3/4x - 5

has positive slope. This leaves D.

6 0
2 years ago
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