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AveGali [126]
3 years ago
7

145 find 2 intergers

Mathematics
1 answer:
Umnica [9.8K]3 years ago
3 0

Answer:

72 and 73

Step-by-step explanation:

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Find the value of y for which line a is parallel to line b.
alexdok [17]
Y+38°= 180 beacuse if a and b are parallel they are co interior angles. This means that they will at to 180.

y=180-38= 142°

Answer: 142°

Drew out some hopefully helpful pictures below. Hope they help! let me know if you still have any questions!

6 0
3 years ago
A store has the following candles in a basket:
Inga [223]
This is your answer.

5 0
3 years ago
A student is asked to factor, if possible, the given expression. The student's response demonstrates which common misconception?
dlinn [17]

Answer:

The student assumed he was asked to factor

(a^2 - b^2)

Which is equal to

= (a+b)*(a-b)    (His response)

But, he was asked to simplify

(a^2 + b^2)

Which has imaginary roots, and can be factored as:

(b +ai)(b-ai)

8 0
3 years ago
Combine the following fractions by addition or subtraction as directed.
Marina86 [1]
Answer:\frac{6}{y^2-xy}-\frac{6}{x^2-xy}=\frac{6(x+y)}{xy(y-x)}

Explanation:

Combining the fractions, we have:

\begin{gathered} \frac{6(x^2-xy)-6(y^2-xy)}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6xy-6y^2+6xy}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6y^2}{x(x-y).y(y-x)} \\  \\ =\frac{6(x-y)(x+y)}{-xy(x-y)^2} \\  \\ =\frac{-6(x+y)}{xy(x-y)} \\  \\ =\frac{6(x+y)}{xy(y-x)} \end{gathered}

8 0
1 year ago
What is the equation of the quadratic function represented by this table?
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

I used logic and took the easy way around this as opposed to the long, drawn-out algebraic way.  I noticed right off that at x = -3 and x = -1 the y values were the same.  In the middle of those two x-values is -2, which is the vertex of the parabola with coordinates (-2, 4).  That's the h and k in the formula I'm going to use.  Then I picked a point from the table to use as my x and y in the formula I'm going to use.  I chose (0, 3) because it's easy.  The formula for a quadratic is

y=a(x-h)^2+k

and I have everything I need to solve for a.  Filling in my h, k, x, and y:

3=a(0-(-2))^2+4  and

3=a(2)^2+4  and

-1 = 4a so

a=-\frac{1}{4}

In work/vertex form the equation for the quadratic is

y=-\frac{1}{4}(x+2)^2+4

In standard form it's:

y=-\frac{1}{4}x^2-x+3

8 0
3 years ago
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