1/6 of a birthday cake is left over from a party. The next day, it is shared among 7 people. How big A piece of the original cake did each person get?
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.
The measures of heights (in inches) of a class of students are shown. 60, 60,54, 62, 62, 57,58, 65, 54,67, 63,57, 62, 54, 72, 66, 54, 69, 55, 61 Find the mean, median, mode, range, and standard deviation of the heights.b. A new student who is 7 feet tall joins your class. How would you expect this student’s height to affect the measures in part (a)?
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:
Exactly 150 years ago, students planted a tree in front of their school. Since then, the tree has grown 21 meters taller and the distance around the trunk has increased by 3 meters. Use pencil and paper. Describe how you can use this information to find at least two different unit rates. Then find these unit rates.
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.
Find the sum of each of the following series:
(a) 2 + 7 + 12 + ...... +92
where u(n) is the nth term (any term).
u⁰ is out first term (2)
d is the common difference (5)
u(n)=u⁰+nd92 =2 +5n5n =90n =18This sequence admits 19 terms since we start counting from 0 till 18.SummationS= number of terms/2 ×(u¹⁸+u⁰)[tex]s = \frac{19}{2} \times (92 + 2)[/tex][tex]s = 9.5(94)[/tex]
[tex]s = 893[/tex]
That's ur answer :)what is the algabric way to find the area of a triangle
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
Area of TriangleParameters for find the area of triangle are
BaseHeightLet 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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A state park is designed in a circular pattern as shown in the diagram. Marta rides her bike along the circular path from the ranger station to the pool. How far does she ride? Round your answer to the nearest tenth of a mile. Show your work please
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
For more information:
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~lenvy~
Isolate the variable to find the solution of the equation. 6. 2w = 18. 6 w = 3 w = 12. 4 w = 24. 8 w = 115. 32.
An equation is formed of two equal expressions. The solution of the equation is w=3.
What is an equation?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.
Learn more about Equation:
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subtract, need help again! thanks a lot.
[tex]\quad\qquad\qquad\huge\underline{\boxed{\sf Aηswer☂ }}[/tex]
Here's the solution ~
# First[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]
# Second[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]
# Third[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]
# Fourth[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]
# Fifth[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]
Find the derivative of following function.
[tex]\\ \rm\rightarrowtail y=\dfrac{cos^2x-3\sqrt{x}+6}{sin^2x+6}\times \dfrac{tan^2x+5x}{cosec^2x+3}[/tex]
Make sure you include proper explanation and all steps .
Wrong answers aren't tolerated.
Spam/short/irrelevant answers will be deleted so don't do time pass if you don't know
[tex]\boxed{\pink{\mathscr{All\:the\:best}}}[/tex]
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
DerivativesDerivative NotationDerivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]
Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]
Derivative Rule [Basic Power Rule]:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative 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
A ball is thrown from an initial height of 4 feet with an initial upward velocity of 33 ft/s. The ball's height h (in feet) after t seconds is given by the following.
h= 4 + 33t - 16t^2
Find all values of t for which the ball's height is 20 feet.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.).
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.
please help me here?
Answer:
$90
Step-by-step explanation:
12 * 4 = 48
48 + (12*3.5) = 90
answer is $90
Keeram ordered 15 1/2 kg of juniper berries for her store. But 2/7 of them arrived bruised and mushy how many kg of useable berries did she receive?
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.
About 11,000 species of Lepidoptera (butterflies and moths) have been identified in the United States. Only 679 of them are butterflies. What is the percentage of the Lepidoptera species in the United States are butterflies?
*Science+Math*
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
10. Jacinto Corado rented a minivan for 4 days in Tampa, Florida, for
$88.00 a day plus $0.20 a mile for all miles over 200. He drove 75, 120,
85, and 140 miles on the days rented. He paid a CDW fee of $18.50 per
day and $61.83 for gasoline.
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 length of a rectangle is 3 meters more than twice its width what is the width of the rectangle if the area is 9 m^2. pls help. solve by using applications of quadratics
Step-by-step explanation:
answer is shown above, that is a tricky question though.
write the standard form of the equation of the circle with the given center and radius. Center (2.-1), r= √3
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
15 points do all pls
A windowpane is 5 inches by 6 inches. What is the distance between opposite corners of the windowpane?
Answer:
6 in.
Step-by-step explanation:
5 inches is the width. Length is 6 in.
You are painting the outside wall of a large building. The wall has a width of 60.11 meters, and a height of 79.88 meters. You need to estimate the total area of the wall, so you know how much paint to buy. Which number below is the best estimate?
A. 4,800 m2
B. 4,400 m2
C. 5,200 m2
D. 5,300 m2
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
A small class has 9 students, 3 of whom are girls and 6 of whom are boys. The teacher is going to choose two of the students at random. What is the probability that the first student chosen will be a boy and the second will be a girl? Write your ansuer as a fraction in simplest form.
What is the measure of an angle at the centre whose radius is equal to arc length?
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.
Hope it helps!!!Brainliest pls!!!The measure of an angle at the center is 1 radian.
What is an angle measure?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.
To learn more about the angle measure;
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#SPJ2
solve pls brainliest
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!
Simplify 6-2a
Thank you
Answer:
4a because first, do 6 - 2 which is 4 and combine with a.
Please mark brainliest, thanks!
Ben and Cam are scuba diving. Ben is 15.8 meters below the surface of the water. Cam is 4.2 meters below
Ben. Andrew is a further 16 meters.
Part 1 Write an expression That'll allow you to find Andrew's depth.
Part 2: What's Andrew's depth?
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.
What is the solution to the
equation log, (2x - 3) = -1?
Round to the nearest thousandth.
[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]
solve pls brainliest
Hi again! The answer is 200 students
Would you like me to help explain so you can do these questions on your own? :)
In which of the following situations should a debit card be used? A Eugene is at a cash-only restaurant, and needs to use the ATM. B Eugene is going to buy a guitar, but he does not have the money for it. с Eugene is not sure how much money is in his account and does not want an overdraft fee. D Eugene was in a car crash, and now he has a lot of expensive medical bills to pay.
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.)
Events that cannot happen at the same time are called?
I need help please provide the steps
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
What heat transfer is Christmas lights radiation or convention or conduction
Answer:
The heat from the sun or heat released from the filament of a light bulb is the example of radiation.