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Dmitrij [34]
3 years ago
9

5.Solve the system using substitution.

Mathematics
1 answer:
svlad2 [7]3 years ago
4 0
Equation 1)  y - 2x = 3
Equation 2)  3x - 2y = 5

Switch around x and y in equation 1.

1)  -2x + y = 3

2)  3x - 2y = 5

Multiply ALL of equation 1 by 2.

1)  2(-2x + y = 3)

Simplify.

1)  -4x + 2y = 6

2)  3x - 2y = 5

Add equations together.

-x = 11

Divide both sides by -1.

x = -11

Plug in -11 for x in the first equation.

-4x + 2y = 6

-4(-11) + 2y = 6

Simplify.

44 + 2y = 6

Subtract 44 from both sides.

2y = 6 - 44

Simplify.

2y = -38

Divide both sides by 2.

y = -38/2

y = -19

So, x = -11, y = -19

D : (-11, -19)

~Hope I helped!~
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Step-by-step explanation:

It seems like it is going down by -30 each term.

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Here is a riddle: “I am thinking of two numbers that add up to 5.678. The difference between them is 9.876. What are the two num
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Step-by-step explanation:

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Arlecino [84]
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6 0
3 years ago
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Instructions: Find the lengths of the other two sides of the isosceles right triangle below.
Alexandra [31]

Given:

The ratio of 45-45-90 triangle is x:x:x\sqrt{2}.

The hypotenuse of the given isosceles right triangle is 7\sqrt{2}.

To find:

The lengths of the other two sides of the given isosceles right triangle.

Solution:

Let l be the lengths of the other two sides of the given isosceles right triangle.

From the given information if is clear that he ratio of equal side and hypotenuse is x:x\sqrt{2}. So,

\dfrac{x}{x\sqrt{2}}=\dfrac{l}{7\sqrt{2}}

\dfrac{1}{\sqrt{2}}=\dfrac{l}{7\sqrt{2}}

\dfrac{7\sqrt{2}}{\sqrt{2}}=l

7=l

Therefore, the lengths of the other two sides of the given isosceles right triangle are 7 units.

6 0
3 years ago
How would I do question 5 I don’t understand
bazaltina [42]

Answer:

a) f(0)=4

b) f(-2)=2

c) x∈{-8,-4,4,8}

d) x∈{-9,-3,3,9}

e) x∈(-∞,-9)∪(-3,3)∪(9,∞).

f)  x∈[-9,-3]∪[3,9]

g) Domain is x∈[-10,10]. Range is y∈[-4,4]. Zeros are at elements of x in {-9,-3,3,9}

h) Increasing on x∈ (-6,0) ∪ (6,10) and

decreasing on x∈ (-10,-6) ∪ (0,6).

Step-by-step explanation:

a) f(0) means what is the y-coordinate of the point at x=0.

So find 0 on the x-axis. I'm going to go up because the curve is at y=4 there.

Conclusion: f(0)=4.

b) f(-2) means what is the y-coordinate of the point at x=-2.

So find -2 on the x-axis. I'm going to go up because the curve is at y=2 there.

Conclusion:  f(-2)=2.

c) f(x)=-2 means what is x when y=-2.  So go on the y-axis and find -2. Now anything on that line y=-2 (horizontal line going through y=-2 on the y-axis) we need to look at.

There are 4 value we need to look at then.

x=-8

x=-4

x=4

x=8

At all of these the y-coordinate is -2.

d) f(x)=0 means what is x when y=0.  So go on the y-axis and find 0. Now anything on that line y=0 (horizontal line going through y=0 on the y-axis; also knowing as the x-axis for y=0) we need to look at.

There are 4 values we need to look at then.

x=-9

x=-3

x=3

x=9

e)  f(x)>0 means where is the curve above the x-axis.

The curve is above the x-axis:

  • Before x=-9
  • Between x=-3 and x=3
  • After x=9

The interval notation is:

(-∞,-9)∪(-3,3)∪(9,∞).

d) f(x)≥0 means where is the curve below or on the x-axis (also known as y=0).

The curve is below or on the x-axis:

  • Between -9 and -3 (inclusive of both endpoints because they include y=0).
  • Between 3 and 9 (inclusive of both endpoints because they include y=0).

The interval notation is:

[-9,-3]∪[3,9]

g) The domain is where the curve exists for the x-values.

The domain is all real numbers between -10 and 10 (inclusive of both).

The curve starts at x=-10 and stops at x=10. The function exists a y value for any number between -10 and 10 (including both).

Interval notation is:

[-10,10]

The range is where the curve exists for the y-values.

Looking from bottom to top I see that it starts at y=-4 and stops at y=4. I notice the curve exists at some point between those two horizontal lines. It also exists at both of those endpoints" -4 and 4.

The range is between -4 and 4 (including both).

Interval notation is:

[-4,4]

The zeros are where the graph crosses the x-axis.

The graph crosses the x-axis at:

x=-9

x=-3

x=3

x=9

h)  Reading left to right the graph increases when you see the curve going up.

I see this from x=-6 to x=0 (exclusive of both).

I see this from x=6 to x=10 (exclusive of both).

So interval notation is (-6,0) ∪ (6,10).

Reading left to right the graph decreases when you see the curve going down.

I see this from x=-10 to x=-6 ( exclusive of both).

I see this from x=0 to x=6 (exclusive of both).

4 0
3 years ago
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