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sergij07 [2.7K]
4 years ago
8

Help please

Mathematics
2 answers:
inn [45]4 years ago
7 0
An easy way to do it is subtituet (-7,5) and (-4,1) into the equations and find out which ones check. remember (x,y)

so A
if we try (-7,5)
5-7=-3/4(-7+5)
we can immediately see that -2 is not equal to -3/4(-2)
so A is wrong


B
if we try (-7,5)
5-5=-4/3(-7+7)
0=-4/3(0)
0=0
so B is correct

C
if we try (-7,5)
5+4=-3/4(-7+1)
9=-3/4(-6)
9 is not equal to 18/4
so C is wrong

D
if we try (-7,5)
5-1=-4/3(-7+4)
4=-4/3(-3)
4=12/3
4=4
D is correct

If you wanted to try the second point (-4,1) in B and D
B
1-5=-4/3(-4+7)
-4=-4/3(3)
-4=-12/3
4=4
B is still correct

D
1-1=-4/3(-4+4)
0=-4/3(0)
0=0
D is still correct

so the answers are B and D
Novosadov [1.4K]4 years ago
5 0
General\ equation\ for\ line\ in\ slope\ intercept\ form:\\\\y=ax+b\\\\To\ find\ a\ and\ b\ substitude\ points\ (-7,5)\ (-4,1)\ into\ equation\\\\ 
\left \{ {{5=-7a+b}\ \ \ \ \atop {1=-4a+b\ |*-1}} \right.\\\\ \left \{ {{5=-7a+b}\ \ \ \ \atop {-1=4a-b}} \right.\\\\
+----\\\\Addition\ method\\\\
4=-3a\ \ |:(-3)\\\\
a=-\frac{4}{3}\\\\
b=5+7a=5+7*(-\frac{4}{3})=5-\frac{28}{3}=-4\frac{1}{3}\\\
Answer\ y=-\frac{4}{3}x-4\frac{1}{3}\\\\
Checking\ solution:\\
y-5=-\frac{4}{3}(x+7)
y-5=-\frac{4}{3}x-\frac{28}{3}\\\\
y=-\frac{4}{3}x-\frac{28}{3}+5\\\\
y=-\frac{4}{3}x-4\frac{1}{3}\\\\Answer:B \ and\ D.
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In order to find the value of y at any given x, simply replace x by that number.

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3 years ago
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As part of quality-control program, 3 light bulbs from each bath of 100 are tested. In how many ways can this test batch be chos
hichkok12 [17]

Answer:

<h3>By 161700 ways this test batch can be chosen.</h3>

Step-by-step explanation:

We are given that total number of bulbs are = 100.

Number of bulbs are tested = 3.

Please note, when order it not important, we apply combination.

Choosing 3 bulbs out of 100 don't need any specific order.

Therefore, applying combination formula for choosing 3 bulbs out of 100 bulbs.

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Plugging n=100 and r=3 in above formula, we get

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Expanding 100! upto 97!, we get

=\frac{100\times 99\times 98\times 97!}{97!3!}

Crossing out common 97! from top and bottom, we get

=\frac{100\times 99\times 98}{3!}

Expanding 3!, we get

=\frac{100\times 99\times 98}{3\times 2\times 1}

= 100 × 33  × 49

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FinnZ [79.3K]

Answer:

correct choice is 1st option

Step-by-step explanation:

Two given triangles have two pairs of congruent sides: one pair of length 7 units and second pair of length 8 units. The third side is common, i.e the lengths of third sides are equal too.

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