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3 years ago

Solve -2 sin2x=2 All real solutions

Mathematics

1 answer:

cupoosta [38]3 years ago

3 0

X=arcsin(-1)/2

I hope this helps.

Have a "AWESOME" day. :)

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Rite an algebraic expression to represent "five plus the quotient of r and 14."

hichkok12 [17]

The answer is 3) 5+r/14

Let's go segment by segment:
<span>"five plus" means addition, so we have 5 +

</span><span>"the quotient of r and 14" is the division result, so sign / must be present: 14/r
</span>
Therefore "five plus the quotient of r and 14" is 5 + 14/r

4 0

3 years ago

The function f(t)= 5 tan 2 t, does not have an amplitude and has a period of π.

trasher [3.6K]

ANSWER

False

EXPLANATION

The tangent function has no amplitude because it is not bounded.

The given tangent function is

$f(t) = 5 \tan(2t)$

This is of the form

f(t)=a tan(bt)

The period is given by

$T = \frac{\pi}{ |b| }$

$T = \frac{\pi}{ |2| } = \frac{\pi}{2}$

The first statement is true but the second is false.

Hence the whole statement is false.

8 0

3 years ago

Read 2 more answers

Write 7^3 × √ 7 as a single power of 7

olga_2 [115]

Step-by-step explanation:

7^3 × 7^(1÷2)

= 7^(3+ 0.5)

=7^3.5 ans

8 0

3 years ago

If the perimeter of a square is 68 feet, find the approximate length of its diagonal.A. 17 feet B. 24 feet C. 38 feet D.

Elanso [62]

Find the side length:

Side = 68/4 = 17

Now use Pythagorean theorem:

Diagonal = sqrt(17^2 + 17^2)

Diagonal = sqrt(289 + 289)

Diagonal = sqrt(578)

Diagonal = 24 feet

The answer is B. 24 feet

8 0

3 years ago

Given: JK tangent, KH=16, HE=12 Find: JK.

Y_Kistochka [10]

Answer: $JK=8$

Step-by-step explanation:

You can observe in the figure that JK is a tangent and KH is a secant and both intersect at the point K. Then, according to the Intersecting secant-tangent Theorem:

$JK^2=KE*KH$