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STALIN [3.7K]
3 years ago
5

Solve for n in the literal equation a(n − 5) + 6 = bn.

Mathematics
2 answers:
svetoff [14.1K]3 years ago
8 0

Answer:

a(n - 5) + 6 = bn \\ an - 5a + 6 = bn \\ an - bn = 5a - 6 \\ n(a - b) = 5a - 6 \\ n =  \frac{5a - 6}{a - b}  \\ thank \: you

Wewaii [24]3 years ago
5 0

Answer:

(6-5a)/(b-a) = n

Step-by-step explanation:

a(n − 5) + 6 = bn

Distribute

an -5a +6 = bn

Subtract an from each side

an -5a+6 -an = bn-an

6-5a = bn-an

Factor out n

6-5a = n(b-a)

Divide each side by (b-a)

(6-5a)/(b-a) = n

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What is 5(2a + 3b) expanded
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2 years ago
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6. what is the parimeter of triangle ABC<br>8. What is the value of x​
nalin [4]

<u>Answer:</u>

Perimeter = 20 units

x = 120°

<u>Step-by-step explanation:</u>

We are given a triangle ABC with known side lengths for all three sides and an inscribed circle.

We are to find the perimeter of triangle ABC and the value of x.

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The kite shape at the end is a quadrilateral which has a sum of angles of 360 degrees.

Two out of four angles are right angles and one is 60 so we can find the value of x.

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3 years ago
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To print tickets, a printer chargers a $70 setup fee plus $1.25 per ticket. (a) Write an algebraic expression for the cost of t
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8 0
3 years ago
Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec
german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

#3

Let's start with sin here. \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}} Therefore, because sin is y/r, r = \sqrt{5} and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = \frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}

#4

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = 4\sqrt{2}. Now, to find csc, we have to do r/y.

csc = \frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

#5

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = \sqrt{7/16} = \frac{\sqrt{7}}{4}

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = -\sqrt{7/9} = \frac{-\sqrt{7}}{3}

#6

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = (\frac{-\sqrt{6}}{2})^{2}

1 + cot^2 = 6/4

cot^2 = 2/4

cot = \frac{\sqrt{2}}{2}

tan = \sqrt{2}

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

(\frac{2}{\sqrt{2}})^2 + 1 = sec^2

4/2 + 1 = sec^2

2 + 1 = sec^2

sec^2 = 3

sec = -\sqrt{3}

cos = \frac{-\sqrt{3}}{3}

Hope this helps!! :)

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2 years ago
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