$-7x-3x+2=-8x-8\\\\(-7x-3x)+2=-8x-8\\\\-10x+2=-8x-8\qquad|\text{subtract 2 from both sides}\\\\-10x=-8x-10\qquad|\text{add 8x to both sides}\\\\-2x=-10\qquad|\text{divide both sides by (-2)}\\\\\boxed{x=5}$

8 0

3 years ago

Read 2 more answers

14. In a right triangle, if one acute angle has measure 10x + 6 and the other has measure 2x + 7, find the larger of these two a

lubasha [3.4K]

Answer:

421/6 degrees

Step-by-step explanation:

The acute angles of a right triangle are complementary, so

$10x+6+2x+7=90 \\ \\ 12x+13=90 \\ \\ 12x=77 \\ \\ x=\frac{77}{12}$

So, the acute angles measures 421/6 degrees and 119/6 degrees, and thus the larger angle is 421/6 degrees.

7 0

2 years ago

Dont really understand how to do this

bija089 [108]

Answers:

$t_{10} = -22 \ \text{ and } S_{10} = -85$

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Explanation:

$t_1 = \text{first term} = 5\\t_2 = \text{first term}-3 = t_1 - 3 = 5-3 = 2$

Note we subtract 3 off the previous term (t1) to get the next term (t2). Each new successive term is found this way

$t_3 = t_2 - 3 = 2-3 = -1\\t_4 = t_3 - 3 = -1-3 = -4$

and so on. This process may take a while to reach $t_{10}$

There's a shortcut. The nth term of any arithmetic sequence is

$t_n = t_1+d(n-1)$

We plug in $t_1 = 5 \text{ and } d = -3$ and simplify

$t_n = t_1+d(n-1)\\t_n = 5+(-3)(n-1)\\t_n = 5-3n+3\\t_n = -3n+8$

Then we can plug in various positive whole numbers for n to find the corresponding $t_n$ value. For example, plug in n = 2

$t_n = -3n+8\\t_2 = -3*2+8\\t_2 = -6+8\\t_2 = 2$

which matches with the second term we found earlier. And,

$tn = -3n+8\\t_{10} = -3*10+8\\t_{10} = -30+8\\t_{10} = \boldsymbol{-22} \ \textbf{ is the tenth term}$

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The notation $S_{10}$ refers to the sum of the first ten terms $t_1, t_2, \ldots, t_9, t_{10}$

We could use either the long way or the shortcut above to find all $t_1$ through $t_{10}$ . Then add those values up. Or we can take this shortcut below.

$Sn = \text{sum of the first n terms of an arithmetic sequence}\\S_n = (n/2)*(t_1+t_n)\\S_{10} = (10/2)*(t_1+t_{10})\\S_{10} = (10/2)*(5-22)\\S_{10} = 5*(-17)\\\boldsymbol{S_{10} = -85}$

The sum of the first ten terms is -85

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As a check for $S_{10}$ , here are the first ten terms:

t1 = 5
t2 = 2
t3 = -1
t4 = -4
t5 = -7
t6 = -10
t7 = -13
t8 = -16
t9 = -19
t10 = -22

Then adding said terms gets us...

5 + 2 + (-1) + (-4) + (-7) + (-10) + (-13) + (-16) + (-19) + (-22) = -85

This confirms that $S_{10} = -85$ is correct.

6 0

2 years ago

Express sin A, cos A, and tan A as ratios

vlabodo [156]

<h3>Answer: Choice B</h3><h3>sqrt(3)/2, 1/2, sqrt(3)</h3>

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Explanation:

Sine of an angle is the ratio of the opposite side over the hypotenuse. For reference angle A, the opposite side is BC = 6sqrt(3). The hypotenuse is the longest side AB = 12

Sin(angle) = opposite/hypotenuse

sin(A) = BC/AB

sin(A) = 6sqrt(3)/12

sin(A) = sqrt(3)/2

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Cosine is the ratio of the adjacent and hypotenuse

cos(angle) = adjacent/hypotenuse

cos(A) = AC/AB

cos(A) = 6/12

cos(A) = 1/2

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Tangent is the ratio of the opposite and adjacent

tan(angle) = opposite/adjacent

tan(A) = BC/AC

tan(A) = 6sqrt(3)/6

tan(A) = sqrt(3)

5 0

3 years ago