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2 years ago

5

Solve for n. n+ 1 = 4(n-8) n = 1 n = 8 n = 11 C n = 16 Help? ;-;

1 answer:

kari74 [83]2 years ago

8 0

Answer:

MMMMMMMMMMMMMMMM NO :)

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Solving systems by Substitution Homework y=x+4 y=x-3

zvonat [6]

There is no solution for this equation since they have the same slope, x/1. The only difference between these two is that their y intercepts are different meaning that they will be parallel lines that will never intersect among one other. For example, think of it as two separate lines that are have the same slope and never gain more distance/units from one another.

Solve:

To solve, you have to get one of this equations into a Ax+By=C equation form, standard equation. Let’s change y=x+4 into a standard equation.

We have to get x and y together and 4 as C.
So let’s subtract x from both sides;

y=x+4
-x -x
————————
-x+y=4

This is a standard equation.

Now let’s substitute.

take the standard equation and plug in y which is x+4 since there is a equation meaning it’s y=x+4

-x+(x+4)=4

Let’s simplify this mess.

-x+x equals 0. So we are left with 4=4.

Subtract 4 from both sides and we get 0=0

This means there is no solution. Hoped this helped.

8 0

2 years ago

Please help me out :P

Veseljchak [2.6K]

cos θ = $\frac{-4\sqrt{65} }{65}$ , sin θ = $\frac{-7\sqrt{65} }{65}$ , cot θ = 4/7, sec θ = $\frac{-\sqrt{65} }{4}$ , cosec θ = $\frac{-\sqrt{65} }{7}$

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = $\sqrt{7^{2} + 4^{2} }$ = $\sqrt{49+16}$ = $\sqrt{65}$

sin θ = opposite/ hyp = $\frac{7}{\sqrt{65} }$

Rationalize, $\frac{7}{\sqrt{65} }$ x $\frac{\sqrt{65} }{\sqrt{65} }$ = $\frac{7\sqrt{65} }{65}$

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - $\frac{7\sqrt{65} }{65}$

cos θ = adj/hyp = $\frac{4}{\sqrt{65} }$

By rationalizing and knowing that cos θ is negative, cos θ = - $\frac{-4\sqrt{65} }{65}$

cot θ = 1/tan θ = 1/7/4 = 4/7

sec θ = 1/cos θ = 1/ $\frac{4}{\sqrt{65} }$ = - $\frac{-\sqrt{65} }{4}$

cosec θ = 1/sin θ = 1/ $\frac{\sqrt{65} }{7}$ = $\frac{-\sqrt{65} }{7}$

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

5 0

1 year ago

You play a game that involves rolling a die. You can roll as many times as you want, but if you roll a six your score is zero an

anyanavicka [17]

Answer:

6

Step-by-step explanation:

P(6) = ⅙

Expected no. of trials for first 6:

1 ÷ (1/6) = 6

5 0

2 years ago

A quiz has 5 true-false questions and 3 multiple choice questions with 4 options each. How many possible ways are there to answe

8_murik_8 [283]

This answer is d 2024

4 0

2 years ago

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A spherical fish bowl is half-filled with water. T

olganol [36]

The volume of a sphere of radius r is given by

$V=\dfrac{4}{3}\pi\cdot r^3$

For your given conditions, the volume of 1/2 a sphere becomes

$V=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{22}{7}\cdot 8^3$

Seems to match your 2nd choice:

... 1 over 24 over 322 over 7(83)

7 0

2 years ago

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