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aleksandr82 [10.1K]
3 years ago
15

If you know the value of X and Y.. please let me know.​

Mathematics
2 answers:
Natali5045456 [20]3 years ago
6 0
Answer: X= 35° and y= 20°
Hopefully, this was helpful
Fantom [35]3 years ago
5 0
If i’m not wrong the shape is trapezium, the side should be the same which mean it’ll be 65 ay the y side and 125 at x side. so we all can see the first triangle is 90 which we can get x with 125-90=35 so the answer for x=35 then to find x we just need minus the whole sec triangle which is 180-125-35=20 so y=20

hope you understand what i’m trying ti explain for you
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tester [92]

Answer:

x + 29 = 41

Step-by-step explanation:

Trevon received some money, but we don't know how much, so we will represent that with x.

The money he received will be add the amount he had last friday, $29.

The equation will equal his current amount, $41.

6 0
3 years ago
How do you find the x and y intercept with just the slope(y=mx+b)
solong [7]

Answer:

When you have a equation you can easily find the x value and y value which you can assume the x-intercept is (x,0). and y-intercept is (0,y)

Hope this helped.

4 0
3 years ago
Read 2 more answers
Evaluate the expression when x = -4.<br>5(x – 6) + 3x – 2​
SVETLANKA909090 [29]

Answer:

-64

Step-by-step explanation:

5(x – 6) + 3x – 2

Distribute

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Combine like terms

8x -32

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6 0
4 years ago
Find the equation of the graphed line.?
kiruha [24]

Answer:

c.  y = x - 6.

Step-by-step explanation:

It's a straight line whose general form is y = mx + c.

The slope is (0, -6) / (6, 0)

= -6 / -6

= 1 so the coefficient of x , (m)  is 1.

The value of  c ( the y-intercept) is -6.

7 0
3 years ago
The sum of 30 times (1/3)^(n-1) from 1 to infinity
sergeinik [125]

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}

and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

8 0
3 years ago
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