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umka21 [38]
3 years ago
9

How many solutions does 6a- 3a - 6 = 2 + 3 have?

Mathematics
1 answer:
lana [24]3 years ago
4 0
Just one. the solution is 11/3
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We have a system of equations in two variables, namely, a and c. Use the substitution to find a and c.

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I need urgent help this in for my final grade
elena-14-01-66 [18.8K]
What is the question?
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3 years ago
Find the exact value by using a half-angle identity. . cos(5π/12).
vodka [1.7K]
<span>First of all in order to find the exact value by using a half angle identity we will do
cos(5π divided by 12)

=cos (5π divided by 6) divided by 2
= ± Under root[(1+cos (5π divided by 6) divided by 2]
Now we will consider + sign because 5π divided by 12 is in quadrant. where cos>0
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6 0
3 years ago
Given that rectangle LMNO with coordinates L(0,0), M(3,0), N(3,7), O(0,7), P is the midpoint of LM⎯⎯⎯, and Q is the midpoint of
Elina [12.6K]

The midpoint of a line divides the line into equal segments.

The option that proves PQ = LO is (a)

The given parameters are:

\mathbf{L = (0,0)}

\mathbf{M = (3,0)}

\mathbf{N = (3,7)}

\mathbf{O = (0,7)}

P is the midpoint of LM.

So, we have:

\mathbf{P = \frac{LM}{2}}

\mathbf{P = (\frac{(0 +3}{2},\frac{0+0}{2})}

\mathbf{P = (\frac{3}{2},0)}

Q is the midpoint of NO.

So, we have:

\mathbf{Q = \frac{NO}{2}}

\mathbf{Q = (\frac{(3 +0}{2},\frac{7+7}{2})}

\mathbf{Q = (\frac{3}{2},7)}

Distance PQ is calculated as follows:

\mathbf{d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}

This gives:

\mathbf{PQ = \sqrt{(3/2 - 3/2)^2 + (0 - 7)^2}}

\mathbf{PQ = \sqrt{ 7^2}}

\mathbf{PQ = 7}

Distance LO is calculated as follows:

\mathbf{LO = \sqrt{(0 - 0)^2 + (0 - 7)^2}}

\mathbf{LO = \sqrt{ 7^2}}

\mathbf{LO=7}

So, we have:

\mathbf{PQ = 7}

\mathbf{LO=7}

Thus:

\mathbf{PQ = LO}

Hence, the correct option is (a)

Read more about distance and midpoints at:

brainly.com/question/11231122

8 0
2 years ago
X to the third plus x minus 7 equals -3 square root of x - 1
Ugo [173]

Answer:

  x ≈ 1.5004

Step-by-step explanation:

We suppose your equation is ...

  x^3+x-7=-3\sqrt{x-1}

Squaring both sides and subtracting the right side gives ...

  x^6 +2x^4 -14x^3 +x^2 -23x +58 = 0

This 6th-degree equation has two positive real roots, near x=1.5, and x=2. The root at x=2 is extraneous. The one near x=1.5 is irrational.

5 0
3 years ago
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