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Anit [1.1K]
3 years ago
6

Please answer this in two minutes

Mathematics
1 answer:
alexandr1967 [171]3 years ago
3 0

Answer:

u = \sqrt{6}.

Step-by-step explanation:

This is a 45-45-90 triangle.

That means that there are two side lengths with lengths of x, and a hypotenuse with a length of xsqrt(2). We can then set up a proportion.

\frac{1}{\sqrt{3} } =\frac{\sqrt{2} }{u}

1 * u = \sqrt{3} * \sqrt{2}

u = \sqrt{6}.

Hope this helps!

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Write a rule for the nth term of the arithmetic sequence. Then graph the first six terms of the sequence. a6=-12 a12=-36
denis-greek [22]

Answer:

We need to find the first term.  We can use the formula

an = a1 + d(n - 1)

to solve for a1.  We already know the 12th term, a12, common difference, d, and nth sequence, n.

a12 = -36

d = -4

n = 12

-36 = a1 - 4(12 - 1)

-36 = a1 - 44

8 = a1

The first term is 8.  Therefore, your formula is

an = 8 - 4(n - 1)

an = -4n + 12

Then use this formula to graph.

n is the independent variable.

an is the dependent variable.

Your graph will be a line.

n    |       an

___________

1          8

2          4

3          0

4         -4

5         -8

6         -12

Step-by-step explanation:

give me brainliest.

3 0
3 years ago
WHICH IS NOT A LINEAR FUNCTION ​
Tom [10]

Answer:

OPTION C

Step-by-step explanation:

IN OPT. A- 2 IS MULTIPLED IN EACH NO.

IN OPT.B-2 IS MULTIPLED

IN OPT.D-1 IS ADDED

IN OPT.E-2 IS ADDED

4 0
2 years ago
Determine the zeroes of the polynomial
Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)}  \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)}  -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\

\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

8 0
3 years ago
Which graph represents the inequality y>1-3x
slava [35]

Answer:

b............ .......... which one is a b c d

For this case we must indicate the graph of the following inequality:

y≥1−3x

It is observed that inequality includes equality, so the boundary line of the graph will not be dotted, so we discard options D and C.

We test option A, we substitute the point (0,0) in the inequality, if it is fulfilled then the graph corresponds to it.

We test option A, we substitute the point (0,0) in the inequality, if it is fulfilled then the graph corresponds to it.

It is not fulfilled

We test the last option B, we choose the point (3,1) that belongs to the graph:

1≥1−3(3)

\1≥1−9

1≥−8

it is fulfilled

5 0
2 years ago
PLEASE HELP ME PLEASE
fiasKO [112]

Answer:

4/3

Step-by-step explanation:

6 0
3 years ago
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