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2 years ago

Find the value of x. Round to the nearest degree.

Mathematics

1 answer:

Fantom [35]2 years ago

4 0

Answer:

x = 51

Step-by-step explanation:

Here, we want to find the value of x

To do this, we are going to use the appropriate trigonometric identity

We have the side facing the right angle ( hypotenuse) and the side facing the angle given (opposite)

The trigonometric identity that links both is the sine and it is the ratio of the opposite to the hypotenuse

Thus, we have it that;

sine x = 7/9

x = arisine (7/9)

x = 51

You might be interested in

A computer online service charges one hourly rate for regular use but a higher hourly rate for designated "premium" areas. One c

N76 [4]

Answer: the service charge per hour for premium services is $5.5

the service charge per hour for regular services is $3

Step-by-step explanation:

Let x represent the service charge per hour for premium services.

Let y represent the service charge per hour for regular services.

One customer was charged $38 after spending 2 h in premium areas and 9 regular hours. It means that

2x + 9y = 38- - - - - - - - - - - 1

Another customer spent 3 h in premium areas and 6 regular hours and was charged $34.50. It means that

3x + 6y = 34.5- - - - - - - - - - -2

We would eliminate x by multiplying equation 1 by 3 and equation 2 by 2. It becomes

6x + 27y = 114

6x + 12y = 69

Subtracting, it becomes

15y = 45

y = 45/15

y = 3

Substituting y = 3 into equation 1, it becomes

2x + 9 × 3 = 38

2x + 27 = 38

2x = 38 - 27 = 11

x = 11/2 = 5.5

8 0

2 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago

What is 9 copies of 3.65

Nataly_w [17]

I believe that you are doing multiplication with this.

So… 9 x 3.65 = 32.85

Hope this helps. :)

3 0

3 years ago

How do you solve this?

masha68 [24]

Here's what i found online:

a/3+4=6

We simplify the equation to the form, which is simple to understand
a/3+4=6

Simplifying:
 + 0.333333333333a+4=6

We move all terms containing a to the left and all other terms to the right.
 + 0.333333333333a=+6-4

We simplify left and right side of the equation.
 + 0.333333333333a=+2

We divide both sides of the equation by 0.333333333333 to get a.
a=6

Hope this helped :-)

4 0

3 years ago

Manny worked as a produce manager for

Monica [59]

Answer: 64%

Step-by-step explanation:

You would divide 50 by 32 (32/50), and get the decimal 0.64. Multiply 0.64 by 100 and you will get 64, your percent.

3 0

3 years ago