All categories

2 years ago

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

Mathematics

1 answer:

dusya [7]2 years ago

5 0

Answer:

x = 10.2

Step-by-step explanation:

Here we have a right triangle with hypotenuse 11, angle 22° and adjacent side x. Our task is to find x to the nearest tenth.

The cosine function nicely connects the hypotenuse and the adjacent side:

adjacent side

cos 22° = ------------------------

hypotenuse

We solve for the adjacent side (x) by multiplying both sides of this equation by "hypotenuse:"

x = adjacent side = 11*cos 22 = 11(0.9272 radian).

Performing the indicated multiplication, we arrive at:

x = 10.1990, or, rounded to the nearest tenth, x = 10.2.

You might be interested in

From a random sample of 40 people at a dinner party, 15 ordered the steak dinner. If there are a total of 212 people at the dinn

vesna_86 [32]

B maybe or A
Because 15 percent of 212 is 31.8 so it’s closer to 40 maybe

5 0

2 years ago

Rearrange 3x - 6y = -2 to the form y=mx + c

wel

3x - 6y = -2

-6y = -3x - 2, divide through by -1

6y = 3x + 2, divide through by 6

y = 3x/6 + 2/6

y = x/2 + 1/3

5 0

2 years ago

Read 2 more answers

Max is making a rectangular garden that is 5 feet less than twice it's width. If the perimeter of the garden is 80 feet, what wi

EleoNora [17]

15 feet x 25 feet x 15 feet x 25 feet

4 0

2 years ago

If a person lnvests $120 at 8% annual interest, find the approximate value of the investment at the end of 5 years

Pie

Answer:

$176.3193692 or $176.32 (rounded to two decimal places)

Step-by-step explanation:

It is a compound interest, which means the interest accumulates on an initial amount each period.

The formula is A=P(1+R)^n

A= the total amount P=initial amount R=rate n=time (years)

P=$120 Rate= 8% or 0.08 (decimal) n=5 (years)

A=120 (1+0.08)^2

A=120 (1.469328077)

A= 176.3193692

6 0

1 year ago

I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have

$i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079$

3 0

3 years ago