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2 years ago

Find one positive and one negative coterminal angle to 2.7 Radians:

Mathematics

1 answer:

Pavlova-9 [17]2 years ago

4 0

the two coterminal angles to the given one are:

8.98 rad and -3.58 rad

<h3>How to find coterminal angles?</h3>

For a given angle A in radians, the family of coterminal angles is defined as:

B = A + n*(2*pi)

Where pi = 3.14 rad

Where n can be any integer number.

In this case, we have an angle of 2.7 radians, then the coterminal angles are:

B = 2.7 rad + n*(6.28 rad)

One positive is what we get if we select n = 1, then:

B = 2.7 rad + 1*(6.28 rad) = 8.98 rad.

And if we select n = -1, we get the negative coterminal angle:

B' = 2.7 rad - 1*(6.28 rad) = -3.58 rad

Then the two coterminal angles to the given one are:

8.98 rad and -3.58 rad

If you want to learn more about coterminal angles:

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PLEASE HELP What is the rule for the sequence with the first four terms below?

ivolga24 [154]

Answer:

A) [First try is correct for all U1 to U4]

Step-by-step explanation:

1. Trying each option

(let x = 1,2,3,4) since it goes from U1 to U4 (Geometric sequence)

A.) 10(-4/5)^x

10(-4/5)^1 = -8

10(-4/5)^2 = 6.4

10(-4/5)^3 = -5.12

10(-4/5)^4 = 4.096

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3 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

3 years ago

Please answer this for me correctly. THANK YOU!!

Leona [35]

Answer:

Hope this helps!

5 0

3 years ago

What is the anwer of 40+678

dsp73

Answer:

718

Step-by-step explanation:

just add the two numbers

6 0

3 years ago

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Can soembody help me answer this please. PLEASE BE QUICK AND RIGHT !!!

myrzilka [38]

Answer:

A, D, and G

Step-by-step explanation:

Let's try each of the answer choices, one by one.

A) Plug x = 2 into each expression and see what we get.

7x + 4 = 7 * 2 + 4 = 14 + 4 = 18

3x + 5 + 4x - 1 = 3 * 2 + 5 + 4 * 2 - 1 = 6 + 5 + 8 - 1 = 18

Since they're both 18 when x = 2, A is true.

B) The statement says that the expressions are <em>only </em>equal for x = 4 and x = 8, but as we saw from A, they're also equal when x = 2, so B is wrong.

C) Let's try to find a counterexample for this by plugging in an odd number into each expression, like 1:

7x + 4 = 7 * 1 + 4 = 7 + 4 = 11

3x + 5 + 4x - 1 = 3 * 1 + 5 + 4 * 1 - 1 = 3 + 5 + 4 - 1 = 11

We see that these two expressions are equivalent when x = 1, which is an odd number, so this contradicts the statement that they're <em>only</em> equal when x is even. C is wrong.

D) Let's combine like terms for the expression 3x + 5 + 4x - 1:

3x + 4x + 5 - 1 = 7x + 4

Notice that this is the exact same expression as the first one, which is 7x + 4 so that means for any x, these two expressions will always be the same. D is thus true.

E) This is not true because it's possible that two expressions aren't equivalent but have the same value for a certain odd number and a certain even number. E is incorrect.

F) Obviously, if the two expressions are equivalent, they will give the same answer for any input of x, including when x = 0, so F is wrong.

G) Again, the two expressions will always be equal, including when x = 8, so this statement is true.

The answers are A, D, and G.

Hope this helps!

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3 years ago

Read 2 more answers