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natulia [17]
3 years ago
9

Jane says that she can use addition to show that the graphs of 2x - 3y = 1 and 2x + 3y = 2 are intersecting lines. In two or mor

e complete sentences, describe the process of using addition to show that the lines are intersecting.
Mathematics
1 answer:
iogann1982 [59]3 years ago
4 0
For this case we have the following equations:
 2x - 3y = 1
 2x + 3y = 2
 When adding both equations we observe that:
 Sentence 1: we have an equation of a variable, which in this case will be x. We can clear the value of x.
 Sentence 2: Knowing the value of x, we can substitute in any equation and find the value of y.
 Sentence 3: The value of x and y represents the point of intersection of both lines (x, y).
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What is the area of the parallelogram?
Anarel [89]

Answer:

Step-by-step explanation:

The height is connected to the base of 16cm

Area = b * h

b = 16

h = 5

Area = b * h

Area = 16 cm * 5 cm

Area = 80 cm^2

4 0
3 years ago
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which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2
Wewaii [24]

You know that 100=10^2.

Use the property (ab)^2=a^2\cdot b^2 to state that

100n^2=10^2\cdot n^2=(10n)^2.

Then the expression 100n^2-1 is equivalent to

(10n)^2-1=(10n)^2-1^2.

Answer: correct choice is A.

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3 years ago
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Which figures can have all of the properties shown below?
mr_godi [17]

Answer:the 1 or 4

Step-by-step explanation:

6 0
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What is the first significant figure of 13822?
deff fn [24]

Answer:

1

Step-by-step explanation:

The rules for significant figures are:

  • Non-zero digits are always significant.
  • Zeros between significant digits are also significant.
  • Trailing zeros are significant only after a decimal point.

In 13822, all of the digits are non-zero.  So the first significant figure is 1.

3 0
3 years ago
The following rational equation has denominators that contain variables. for this equation, A. write the value or values of the
dsp73

Answer:

A. -4 and 4

B. No solution.

Step-by-step explanation:

The given equation is

\dfrac{3}{x+4}+\dfrac{2}{x-4}=\dfrac{16}{(x+4)(x-4)}

A.

Equate the denominators equal to 0 to find the restrictions on the variable.

x+4=0\Rightarrow x=-4

x-4=0\Rightarrow x=4

Therefore, x\neq -4,4.

B.

We have,

\dfrac{3}{x+4}+\dfrac{2}{x-4}=\dfrac{16}{(x+4)(x-4)}

\dfrac{3(x-4)+2(x+4)}{(x+4)(x-4)}=\dfrac{16}{(x+4)(x-4)}

Multiply both sides by (x-4)(x+4).

3x-12+2x+8=16

5x-4=16

Add 4 on both sides.

5x=16+4

5x=20

Divide both sides by 5.

x=4

Here the solution is x=4 but it is the restricted value.

Therefore, the given equation has no solution.

8 0
3 years ago
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