what’s the answer to this

[tex]~~~~~\sin 30^{\circ} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse} }\\\\\\\implies \dfrac 12 = \dfrac{8}x\\\\\\\implies x = 8\cdot2\\\\\\\implies x = 16 ~ \text{cm.}[/tex]

The measure of the side length is 16 cm after applying the sin ratio which is the side opposite to the angle/hypotenuse.

It is a triangle in which one of the angles is 90 degrees and the other two are sharp angles. The sides of a right-angled triangle are known as the hypotenuse, perpendicular, and base.

It is given that:

A right-angle triangle is shown in the picture.

To find the side length of the hypotenuse apply the sin ratio

sin30 = side opposite to the angle/hypotenuse

1/2 = 8/x

x = 8(2)

x = 16 cm

Thus, the measure of the side length is 16 cm after applying the sin ratio which is the side opposite to the angle/hypotenuse.

Learn more about the right-angle triangle here:

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Answer:

[tex]f(x)=2^x[/tex]

Step-by-step explanation:

Exponential function:

[tex]f(x)=a(b^x)+c[/tex]

Step 1. Find a, y-intercept

Step 2. Find b, slope

Step 3. Find c, horizontal asymptote

a = 1

m = (y₂ - y₁)/(x₂ - x₁)

m = (3 - 1)/(1 - 0) = 2/1

b = 2

c = 0

[tex]f(x)=1(2^x)+0\\f(x)=2^x[/tex]

Answer:  [tex]92 \text{ yd}^2[/tex]

=======================================================

Work Shown:

[tex]A = \frac{h(b_1+b_2)}{2}\\\\A = \frac{8(19+4)}{2}\\\\A = \frac{8(23)}{2}\\\\A = \frac{184}{2}\\\\A = 92\\\\[/tex]

The [tex]b_1[/tex] and [tex]b_2[/tex] are the parallel bases, while h is the height perpendicular to each base.

The units are in square yards, abbreviated as [tex]\text{yd}^2[/tex]

Think of a 1 yard by 1 yard square that has an area of [tex](1 \text{ yd})*(1 \text{ yd}) = (1)^2(\text{yd})^2 = 1 \text{ yd}^2[/tex]

The side of 9 yards is never used and put in there as a distraction.

Answer:

Step-by-step explanation:

Domain is ( - ∞ , ∞ )

Range is [ 0 , 1 ] ∪ ( 2 , ∞ )

By using the definition of the inverse cosine function, we will see that:

[tex]cos^{-1}(0) = pi/2[/tex]

We have the expression:

[tex]cos^{-1}(0) = \frac{pi}{?}[/tex]

Remember that if:

[tex]cos(x) = y\\\\cos^{-1}(y) = x[/tex]

Then, in what do we need to evaluate the cosine function to get an output equal to zero?

[tex]cos( pi/2) = 0[/tex]

Then:

[tex]cos^{-1}(0) = pi/2[/tex]

Then we can replace the question sign with a 2.

If you want to learn more about the cosine function, you can read:

https://brainly.com/question/8120556

Answer:

Two polygons are similar if their corresponding angles are congruent and the corresponding sides have a constant ratio (in other words, if they are proportional). Typically, problems with similar polygons ask for missing sides.

Step-by-step explanation:

this is the correct answer

Answer:

1/216

Step-by-step explanation:

A dice (assuming it is from 1-6) has 6 faces. There are 4 dices.

equation will be

6*6*6=answer

Multiply 6 three times by itself because there are 4 dices. That is because One die would be 1

Two would be 1/6

Three would be 1/36

And four would be the ...

Asnwer=216

Hi!

[tex]A=\pi r^2[/tex]

We will start by solving for the area of the entire figure.

Plugin the values into the circle area formula. We will use 3.14 for pi:

[tex]A=3.14(4)^2[/tex]

Simplify:

[tex]A=50.24[/tex]

Now, this is the area of the entire circle, but we only want that certain shaded region.

We know that a circle has a total of 360 degrees, and the shaded sector is simply 80 degrees.

We can divide to see how much of the circle is shaded:

80 / 360 = 0.22...

Now, we can take this number and multiply it by 50.24 (our answer from above) to get the area of the shaded sector:

0.22 * 50.24 = 11.16

For a similar problem, see:

https://brainly.com/question/11952845

[tex]5 < \sqrt{30} < 6[/tex]

[tex]Locate\ \sqrt{30}\ between\ two\ integers:\\ \downarrow\\ 5 < \sqrt{30} < 6[/tex]

I hope this helps you

:)

Answer:

258.65 mg will be remain after 1000 years

Step-by-step explanation:

This problem can be solved several ways, I prefer the natural log approach for reasons I won’t go into here:

Activity final = Activity initial e^-kt, where

k = the decay constant, = ln2/half life = 0.693/1590, = 0.000436/year

t = the decay time, in the same time base as k

The problem can be worked in units of activity, mass, or number of atoms.

400 x e^-0.000436 x 1000 = 400 x e^-.436 = 258.65 mg

you can check this by estimating. There is about 2/3 of a half life, so there should be more than 200 mg, which would be the value at one half life.

Answer:

About 53022 bacteria present.

Step-by-step explanation:

We can use the continuous growth formula:

[tex]\displaystyle P = A e^{kt}[/tex]

Where k is some constant.

With an initial population of 5000, it grew to 6500 after two hours.

In other words, P = 6500 when A = 5000 and t = 2:

[tex]\displaystyle (6500) = (5000)e^{k(2)}[/tex]

Solve for k:

[tex]\displaystyle \begin{aligned} 6500 & = 5000e^{2k} \\ \\ e^{2k} & = \frac{6500}{5000} = \frac{13}{10} \\ \\ \ln\left(e^{2k}\right) & = \ln\left(\frac{13}{10}\right) \\ \\ 2k & = \ln\frac{13}{10} \\ \\ k & = \frac{1}{2}\ln\frac{13}{10}\end{aligned}[/tex]

Therefore, our equation is:

[tex]\displaystyle P = \bigg{5000e}^{\dfrac{1}{2}\ln\dfrac{13}{10}t[/tex]

After 18 hours, t = 18. Hence:

[tex]\displaystyle \begin{aligned} P(18) & = \bigg{5000e}^{\dfrac{1}{2}\ln\dfrac{13}{10}(18)} \\ \\ & \approx 53022\end{aligned}[/tex]

Therefore, after 18 hours, there will be about 53022 bacteria present.

Answer:

she has about $3750 in her acount.

Step-by-step explanation:

hope this helps you!

Answer:

DWest I think

Step-by-step explanation:

Someone else asked this and this is what they think and it seems right

Answer:

20

Step-by-step explanation:

[tex]\frac{32}{1.6} =\frac{32}{\frac{16}{10}} =32*\frac{10}{16} =\frac{320}{16} =20[/tex]

So the answer is 20

Answer: See below

Step-by-step explanation:

To find the factor,

f(x) = 0

[tex]\begin{aligned}&2 x^{4}-72 x^{2}=0 \\&2 x^{2}\left(x^{2}-36\right)=0 \\&x^{2}\left(x^{2}-36\right)=0 \\&x \times x \times(x-6) \times(x+6)=0 \\&\mathrm{x}=0,0,6,-6\end{aligned}[/tex]

So,

x = 0 multiplicity 2

x = -6 multiplicity 1

x = 6 multiplicity 1

For an even multiplicty, the graph touches the x-axis, and for an odd multiplicty, the graph crosses the x-axis

Therefore,

x = 0 multiplicty 2, Touch

x = -6 multiplicty 1, Cross

x = 6 multiplicty 1, Cross

Answer: I believe that by helping or answering others you can get get questions, but I do know there is a limited amount of questions being able to ask before paying monthly. Hope this helps

Answer:

What you have to do is answer peoples question to gain more

Step-by-step explanation:

To create the dot plot you will list the data in order and use the given numbers to create the box plot.

8.10, 8.20, 8.30, 8.32, 8.34, 8.50, 8.50, 8.50, 8.55, 8.60, 8.60, 8.65, 8.65, 8.90

Lower extreme: 8.10

Upper extreme: 8.90

Median: 8.50

1st/lower quartile: 8.32

3rd/upper quartile: 8.60

Please see the attached picture for the boxplot.

Answer:

30.67

Step-by-step explanation:

[tex] \frac{9}{ \sin(x) } \: = \frac{16}{ \sin(115) } \\ \\ \frac{9}{ \sin(x) } \: = \frac{16}{ 0.91 } \: \: \\ \\ \frac{9 \times 0.91}{16} = \sin(x) \\ x = 30.67[/tex]

Answer:

almonds: $10.20

cashews: $12.40

Step-by-step explanation:

Answer:

g(3) = 24

Step-by-step explanation:

The question is essentially asking you, "With an input of 3, what is the output of the function?" To find g(3), plug "3" into the "x" variable. Then, simplify as much as you can.

g(x) = x³ - x

g(3) = (3)³ - (3)

g(3) = 27 - 3

g(3) = 24

a. The expression for the tension in the string is T = mg

b. The expression for the tension in the chain is T = 2mg

To solve the problem, we need to know what tension is.

This is the opposing force in a stretched material.

The expression for the tension in the string is T = mg

Since the string supports the weight of the object, the tension in the string equals the weight of the particle.

Let

So, T = W

T = mg

So, the expression for the tension in the string is T = mg

The expression for the tension in the chain is T = 2mg

Since the chain supports both strings, the tension in the chain equals the tension in both strings.

Let

So, T' = T + T

T' = 2T

T' = 2W

T' = 2mg

So, the expression for the tension in the chain is T = 2mg

Learn more about tension here:

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Answer:

38

Step-by-step explanation:

16 - 10 = 6

21 - 16 = 5

26 - 21 = 4

31 - 26 = 5

31 + 6 = 38

The next missing number is 38.

From there, continue the pattern.

The book value at the end of year 2 of the equipment depreciated using the sum-of-the-years’-digits method is $65,560.

Depreciation is a method used to expense the cost of an asset. Book value is the cost of the asset less the accumulated depreciation.

Sum-of-the-year digits = (remaining useful life / sum of the years ) x  (Cost of asset - Salvage value)

Depreciation expense in year 1:

Sum of the years = 1 +2 +3 +4 + 5 + 6 + 7 + 8 + 9 + 10+ 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 210

Undepreciated life of the asset = 20

(20 / 210) x ($79,600 - 4,000) = 7200

Depreciation expense in year 2:

19 / 210 x ($79,600 - 4,000) = 6840

Book value = 79600 - 7200 - 6840 = $65,560

To learn more about depreciation, please check: https://brainly.com/question/25552427

para lang po sa points hehe

Answer:

13 this is the ansewer to this

Answer:

[tex]b. \: 16[/tex]

Step-by-step explanation:

Divide the numbers ⬇

[tex](4 + 2) \times 2 + 4[/tex]

Add the numbers ⬇

[tex]6 \times 2 + 4[/tex]

Multiply the numbers ⬇

[tex]12 + 4[/tex]

Add the numbers ⬇

[tex]12 + 4 = 16[/tex]

OPTION B IS RIGHT :)

hope this helps!! Mark me brainliest please!

Answer: B. 16

Step-by-step explanation:

Hope this helps!