Ava buys 3.5 pounds of strawberries for $2.99 per pound. How much does she spend?

Answer:

The heat from the sun or heat released from the filament of a light bulb is the example of radiation.

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:

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:

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:

the answer is A

Step-by-step explanation:

If Eugene is at a cash-only restaurant he must withdraw cash out of his bank account (linked to debit card.)

Answer:

radian

Step-by-step explanation:

A radian is a unit of angle that is equal to the angle created at the center of a circle whose arc is equal in length to the radius.

The measure of an angle at the center is 1 radian.

When two lines or rays intersect at a single point, an angle is created. The vertex is the term for the shared point. An angle measure in geometry is the length of the angle created by two rays or arms meeting at a common vertex.

Given:

Radius is equal to arc length.

To find the measure of the angle at the center:

We have the formula,

Angle in radian = arc length / radius

Let a be the arc length,

then a be the radius.

Substituting the values,

Angle in radian = a/a

Angle in radian = 1 radian

Therefore, 1 is the measure of an angle at the center.

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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:

j

Step-by-step explanation:

Explanation:

Draw 6 identical rectangles side by side as shown in figure 1 in the diagram below. Shade one of the rectangles to represent 1/6 of the cake left over.

Now draw in horizontal lines (see figure 2) to split the leftover amount into 7 equal portions. Notice that we have a 7 by 6 grid of tiny rectangles. There are 7*6 = 42 tiny rectangles in total.

Each person will therefore get 1/42 of the original cake when splitting up this last leftover slice.

Answer:

6 in.

Step-by-step explanation:

5 inches is the width. Length is 6 in.

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: 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.

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.

Hi again! The answer is 200 students

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

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

Answer:

$90

Step-by-step explanation:

12 * 4 = 48

48 + (12*3.5) = 90

answer is $90

[tex]\qquad\qquad\huge\underline{\boxed{\sf Answer}}[/tex]

Let's solve ~

[tex]\qquad \sf  \dashrightarrow \: log(2x - 3) = - 1[/tex]

[tex]\qquad \sf  \dashrightarrow \:2x - 3 = {10}^{ - 1} [/tex]

[tex]\qquad \sf  \dashrightarrow \:2x = 0.1 + 3[/tex]

[tex]\qquad \sf  \dashrightarrow \:x = 3.1 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:x = 1.55[/tex]

Answer:

At both 0.78 and 1.3 seconds, to the nearest hundredth, the ball will be at 20 feet.  

Step-by-step explanation:

We can use two approaches to finding the answer:  Calculation and Graphing.

Calculation:

Set h to 20 feet and solve the equation using the quadratic formula.

h= 4 + 33t - 16t^2

20= 4 + 33t - 16t^2

- 16t^2 + 33t + 4 = 20

- 16t^2 + 33t + -16 = 0

Solve using the quadratic formula.  I find solutions of 0.779 and 1.28 seconds.

Graphing:

See the attached graph.

The ball is at 20 feet at 0.783 seconds on the way up and 1.28 seconds on the way back down.

===

Both approaches agree:  0.78 and 1.3 seconds, to the nearest hundredth, the ball will be at 20 feet.  

[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]

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

Thus the second option is the correct answer

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:

(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

Answer:

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

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Answer:

2.3 miles

Step-by-step explanation:

Given by the circular pattern as shown in the diagram:

Ranger station to Nature Center = 90°

Nature center to petting zoo= 40°

Petting zoo to Tennis court =120°

Tennis court to Pool = 65°

Pool to Ranger station= 45°

The radius of the circular pattern = 1 mile

Formula:

Arc length = 2πr angle / 360° where r is radius and π=3.14

Solution:

Thus,

Arc length = 2×3.14×1×90°/360°

Arc length=2×3.14×1×1/4 =1.57 miles

Distance from Nature Center to Petting Zoo

Arc length= 2×3.14×1×40°/360°

Arc length= 2×3.14×1×1/9 = 0.70 miles

Total =1.57+0.70 =2.27 miles

Answer to the nearest tenth = 2.3 miles

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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

The parameters for computing the area of triangle is first listed out and the expression for the area of triangle was used to find the algebraic method

Parameters for find the area of triangle are

Let the base be x, and let the height be y

We know that the expression for the area of triangle is given as

Area = 1/2 (Bass* Height)

Substituting our parameters we have

Area = 1/2*x*y

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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: