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adell [148]
2 years ago
5

Pls help quick

Mathematics
2 answers:
IRISSAK [1]2 years ago
8 0
Give the person above brainiest
Pepsi [2]2 years ago
4 0

hope it really helps......!!!!!

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HELP PLEASE!Write the equation of the line parallel to y = –x + 5 passing through the point (9, –1).
Westkost [7]
Y=-x+8
gradient of the parallel line is the same as the gradient of the line given, which is -1
to find the y-intercept, sub the values (9,-1) into the equation we are finding which is y=-x+c
5 0
3 years ago
HELP ME Four students each wrote an equation. Which two students wrote equations that have no solution?
rjkz [21]

Answer:

Beto and Mark

Step-by-step explanation:

When you simplify Beto's equation, you get 0=5 which is inncorrect. 5 does not equal 0 so there are no solutions. When you simplify Mark's equation, you get 6=2 which is also inncorrect. 6 can not equal 2. If you solve the other 2 equations you get answers that can work.

6 0
3 years ago
Read 2 more answers
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Sloan [31]

Answer:

Sum of \frac{7x}{x^2-4} and \frac{2}{x+2} is \mathbf{\frac{9x-4}{x^2-4}}

Option B is correct answer.

Step-by-step explanation:

We need to find sum of \frac{7x}{x^2-4} and \frac{2}{x+2}

Finding sum of  \frac{7}{x^2-4} and \frac{2}{x+2}:

\frac{7x}{x^2-4}+\frac{2}{x+2}

We know that x^2-4 =(x+2)(x-2)

Replacing x^2-4

\frac{7x}{(x+2)(x-2)}+\frac{2}{x+2}

Now, taking LCM of (x+2)(x-2) and (x+2) we get (x+2)(x-2)

=\frac{7x+2(x-2)}{(x+2)(x-2)}\\=\frac{7x+2x-4}{(x+2)(x-2)}\\=\frac{9x-4}{(x+2)(x-2)}\\=\frac{9x-4}{x^2-4}

So, Sum of \frac{7x}{x^2-4} and \frac{2}{x+2} is \mathbf{\frac{9x-4}{x^2-4}}

Option B is correct answer.

6 0
2 years ago
1/4 of women have a disability. If there are 56 women in a
slavikrds [6]

Answer:

14

Step-by-step explanation:

Because 1/4 of women have a disability, you would expect 1/4 * 56 = 14 women to have a disability out of 56 women.

6 0
3 years ago
Read 2 more answers
What is the fewest pairs of corresponding parts that have to be congruent to establish that two triangles are congruent? Once yo
sveticcg [70]

It seems there are always three pairs: SSS, side side side; SAS, side angle side; ASA, angle side angle.

For a right triangle (HL, LL) it seems there are only two items needed, but that's because we already know one angle (it's right).

Once we have congruent triangles, all the corresponding parts are congruent.  That's three sides and and three vertices, so I'd answer six.  Of course almost everything about the triangles is the same, the area, the perimeter, etc.



3 0
3 years ago
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