A teenager puts $50 into an investment account in January. If by December the balance in the account increases by 17%, what is the amount of money in the account in December?

A.$67.00
B.$58.50
C.$41.50
D.$33.00

Answer:Compute the variance: 49+25+16+9+1+16+36+648=27.

The variance is a measure of the dispersion and its value is lower for tightly grouped data than for widely spread data. In the example above, the variance is 27. What does it mean to say that tightly grouped data will have a low variance? You can probably already imagine that the size of the variance also depends on the size of the data itself. Below we see ways that mathematicians have tried to standardize the variance.

Step-by-step explanation:

Answer:

27.6

Step-by-step explanation:

1) Say bens depth is x

Cam's depth is x + 4.2

Andrew's depth is x + 4.2 + 16 or x + 11.8

1) andrew's depth = x + 11.8, where x is 15.8.

Therefore, his depth is 27.6 meters.

Answer:

1/[tex]\sqrt{2}[/tex] - 1/4

Step-by-step explanation:

We can write 2[tex]\sqrt{8}[/tex] as 2*[tex]\sqrt{4}[/tex]*[tex]\sqrt{2}[/tex]

That becomes 4[tex]\sqrt{2}[/tex], since the square root of 4 is 2 (±2, to be exact)

(4/(4[tex]\sqrt{2}[/tex])  - ( [tex]\sqrt{2}[/tex]/4[tex]\sqrt{2}[/tex])

1/[tex]\sqrt{2}[/tex] - 1/4

Answer:

$531.83

Step-by-step explanation:

88*4=352

75+120+85+140=420. Take away the 200 he was allowed. Leaves 220

220*.20 =44

18.50*4=74

gas is set at 61.83

Sum each of those.

352+44+74+61.83= $531.83

Answer:

Distributive property. When you factor out a GCF, from 12x + 6y, you would get 6(2x+y).

Step-by-step explanation:

12x + 6y is equivalent to 2x + y using the distributive property of multiplication over addition.

A number system is defined as a system of writing to express numbers.

The distributive property states that a(b + c) = ab + ac, where a, b, and c are any real numbers.

We can apply this property to the expression 2x + y by multiplying 2x by 6 and y by 6, giving:

12x + 6y = 6(2x) + 6(y)

Then, we can use the distributive property again to simplify the expression inside the parentheses:

6(2x) + 6(y) = 6(2x + y)

Substituting this expression back into the original equation, we get:

12x + 6y = 6(2x + y)

Simplifying further by dividing both sides by 6, we obtain:

2x + y = 12x/6 + 6y/6

2x + y = 2x + y

Therefore, we have shown that 12x + 6y is equivalent to 2x + y using the distributive property of multiplication over addition.

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[tex]\quad\qquad\qquad\huge\underline{\boxed{\sf Aηswer☂ }}[/tex]

Here's the solution ~

[tex]\qquad \sf  \dashrightarrow \: \sqrt{8} - \sqrt{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: ( \sqrt{2 \times 2 \times 2} )- \sqrt{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 2\sqrt{ 2} - \sqrt{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ 2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt[3]{24} - \sqrt[3]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: (\sqrt[3]{2 \times 2 \times 2 \times 3}) - \sqrt[3]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 2\sqrt[3]{3} - \sqrt[3]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt[3]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt[]{125} - \sqrt{45} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt[]{5 \times 5 \times 5} - \sqrt{3 \times 3 \times 5} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 5 \: \sqrt[]{5 } - 3\sqrt{ 5} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 2 \: \sqrt[]{5 } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \: \sqrt[5]{96 } - \sqrt[5]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \: \sqrt[5]{2 {}^{5} \times 3} - \sqrt[5]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \: 2\sqrt[5]{3} - \sqrt[5]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \: \sqrt[5]{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \: \sqrt[]{75} - \sqrt{27} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \: \sqrt[]{3 \times 5 \times 5} - \sqrt{3 \times 3 \times3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 5 \sqrt{3} - 3 \sqrt{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 2 \sqrt{3} [/tex]

Answer:

Step-by-step explanation: 5.6 would be the add on to 27 so since it included the key word '' left'' you'd substract the two and you'd get your answer, just use calculator. You got this!!!

Go sub to clan fbib !!!!!

Answer:

$90

Step-by-step explanation:

12 * 4 = 48

48 + (12*3.5) = 90

answer is $90

Answer:

Below,

Step-by-step explanation:

f(x)=-(x+2)^2+4

(a) Vertex form

(b)  a = -1 , h = -2 and k = 4.

     vertex = (-2, 4)

(c) The value of a, -1 will invert it vertically or we can describe it as a reflection in the line y = 4. The graph will change from a U shape to a ∩ shape.

Answer:

a) 441 gallons

b) -243 fish

Step-by-step explanation:

For a), multiply the 7 gallons of water by the 63 minutes to get 441 total gallons of water added

For b), multiply the 27 fish by 9 days to find the total decrease in fish, which is -243 fish.

Hopefully this helps - let me know if you have any questions!

Answer:

45 divide by 19 =2 r 7 thanks

Answer:

4a because first, do 6 - 2 which is 4 and combine with a.

Please mark brainliest, thanks!

Answer:

11.07~

Step-by-step explanation:

2/7 of 15.5 is 4.42857142857 which we can round to 4.43. We can then just do 15.5-4.43 to get our answer of 11.07.

Answer:

First Blank: “controlled” Second Blank: "a Different" Third Blank: "the same" Last Blank: "Height of the ramp"

Step-by-step explanation:

Answer: There are 2 different unit rates here:

The increase of the tree's height per year

The increase of the tree's trunk per year

Rate of tree's height per year

= 33m / 150 years = 0.22m per year.

Rate of tree's trunk per year

= 3m / 150 years = 0.02m per year.

[tex] \sqrt[3]{ {x}^{5}y } = {x}^{ \frac{5}{3} } {y}^{ \frac{1}{3} } \\ [/tex]

Thus the second option is the correct answer

An equation is formed of two equal expressions. The solution of the equation is w=3.

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

In order to solve the equation, we need to isolate the variable in the following manner, therefore, the solution of the equation can be written as,

[tex]6.2w=18.6[/tex]

Divide both sides of the equation by 6.2,

[tex]6.2w=18.6\\\\\dfrac{6.2w}{6.2}=\dfrac{18.6}{6.2}\\\\w = 3[/tex]

Hence, the solution of the equation is w=3.

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Answer:

the volume should be 9.42

Answer:

66 (no other answer)

Step-by-step explanation:

The number can be divided by 2 so its even. Its a multiple of 11. It is between 20 and 79. it is a multiple of 3.

Answer: A)  4,800 m2

60.11m x 79.88= 4801.5868

Answer:

A. 4,800 m²

Step-by-step explanation:

Given :

=> width of wall is 60.11 meters.

=> height of wall is 79.88 meters.

The area of the wall is = length × breadth

=> 60.11×79.88

=> 4,801.5868m²

Approx. 4,800

Answer:

[tex]\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}[/tex]

General Formulas and Concepts:
Calculus

Differentiation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

Derivative Rule [Product Rule]:
[tex]\displaystyle (uv)' = u'v + uv'[/tex]

Derivative Rule [Quotient Rule]:
[tex]\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}[/tex]

Derivative Rule [Chain Rule]:
[tex]\displaystyle [u(v)]' = u'(v)v'[/tex]

Step-by-step explanation:

*Note:

Since the answering box is way too small for this problem, there will be limited explanation.

Step 1: Define

Identify.

[tex]\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}[/tex]

Step 2: Differentiate

We can differentiate this function with the use of the given derivative rules and properties.

Applying Product Rule:

[tex]\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '[/tex]

Differentiating the first portion using Quotient Rule:

[tex]\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}[/tex]

Apply Derivative Rules and Properties, namely the Chain Rule:

[tex]\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}[/tex]

Differentiating the second portion using Quotient Rule again:

[tex]\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}[/tex]

Apply Derivative Rules and Properties, namely the Chain Rule again:
[tex]\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}[/tex]

Substitute in derivatives:

[tex]\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}[/tex]

Simplify:

[tex]\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}[/tex]

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

[tex]\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}[/tex]

∴ we have found our derivative.

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Topic: Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Hi again! The answer is 200 students

Would you like me to help explain so you can do these questions on your own? :)

Answer:

6.4

Step-by-step explanation:

she has been riding her bike for 6.4 hours.

Answer:

72%

Step-by-step explanation:

We have 10 sudents in class 1 and 40 in the other for a total of 50 students.

The 10 in class 1 had a mean of 80 so we do 80x10=800

The mean of the class of 40 is 70, so we do 40x70=2800

we add both of those together to get 3600 then we divide that by 50 because there are 50 students in total.

3600/50=72.

Answer:

total no. of butterflies=11000

no.of butterflies=679

percentage of butterflies=?

Percentage of butterflies= (11000/679) percent..

= 16.20 percentage

therefore 16.20 percentage butterflies are in US

Answer:

(x-2)^2 +  (y+1)^2 = 3

Step-by-step explanation:

That is the standard equation for a circle given h,k center and r