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Semenov [28]
3 years ago
6

Find the values of x and y

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
7 0

Answer:

7x-37=4x+8

3x=45

x=15

8y-48 = 3y + 52

5y = 100

y=20

Answer: x=15, y=20

:D

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Solve - 3 [ ( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 ] + 3 ( x - 1 ) (and could you please explain it to me?? I really don't get it.
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Answer:

Step-by-step explanation:

(x+2)(x-1)= x^2+2x-x-2= x^2 +x -2\\\\( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 =x^2+x-2-x^2+1=x-1\\\\- 3 [ ( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 ] =-3(x-1)=-3x+3\\\\- 3 [ ( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 ] + 3 ( x - 1 ) =-3x+3+3x-3=6x\\

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4. Simplify the expression:
Natalija [7]

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<h2>A. 2x² + 5x + 1</h2>

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7x^2-6+4x+7-5x^2+x\qquad\text{combine like terms}\\\\=(7x^2-5x^2)+(4x+x)+(-6+7)\\\\=2x^2+5x+1

3 0
3 years ago
Evaluate the dot product of (5,7) and (-8,2)
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6 0
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Bippity boppity pls help me find this functions property
Semmy [17]

Answer:

Period ⇒ 40

Amplitude ⇒ 12

Mid-line ⇒ 32

Step-by-step explanation:

The table is counting by 4's and the period is the amount of space between 2 peaks. In this scenario, we can find the peaks by looking for two of the same highest value (44). We can see that x=40 has a value of 44 while the other is actually not shown because it would be located at x=0. Therefore the period is 40

<u />

The amplitude can be found by using the following:

\frac{1}{2}|max - min|

Our maximum is 44 and our minimum is 20.

\frac{1}{2}|44 - 20|

\frac{1}{2}|24|

\frac{1}{2}(24)

The amplitude is 12

The amplitude is the distance from the peak to the mid-line. To find the mid-line, we can either subtract our amplitude from our maximum value (44) or add our amplitude to our minimum value (20)

44 - 12 = 32

20 + 12 = 32

Therefore our mid-line is y = 32

~Hope this helps!~

4 0
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