MATHEMATICS HIGH SCHOOL

The area of a rectangular park is 3/5 square mile. the length of the park is 7/8 mile. what is the width of the parkshow your work

remember that Area=Length +Width

Answers

Answer 1
Answer:

Width of a rectangular park is equal to \boldsymbol{(24)/(35)} mile.

Area of a rectangle

Area of a rectangle = Length * Width

Area of a rectangular park =(3)/(5) square mile

Length of a rectangular park =(7)/(8) mile

(3)/(5)=(7)/(8)* Width of a rectangular park

So,

Width of a rectangular park =(3)/(5)* (8)/(7)

                                             =\boldsymbol{(24)/(35)}

Find out more information about rectangle here:

brainly.com/question/15019502?referrer=searchResults
Answer 2
Answer: A = L * W

3/5 = 7/8 * W

divide both sides by 7/8

W = 24/35 of 0.6857

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A taxi cab company charges a flat amount of $5.00 plus a rate of $3.00 per mile traveled, what is the total number of miles traveled if the fare was $41.00

Answers

Answer:

Step-by-step explanation:

Amount charged = $5.00

Amount charged for per mile = $3.00

Miles traveled =

Total amount charged = $41.00

Total amount = $3 per mile - $5

41-5 = 35

Miles traveled= $35 ÷ $3

35÷3 = 11.66

Miles traveled = 11/11.66

A system of linear equations is shown below. 2x+4y=103x−y=8 Maria is attempting to prove that by replacing 2x+4y=10 with a different equation it will sometimes produce a new system of equations with the same solution. Maria plans on multiplying 2x+4y=10 by 2 and then adding the results to the equation 3x−y=8 in order to create a new equation. Maria claims that the new equation that she will replace 2x+4y=10 with is 7x+7y=12. Is Maria correct? Why or why not?

Answers

Answer:

No she is wrong.

Explanation:

\sf\n2x+4y=10...........(1)\n3x-y=8...........(2)\n\n\textsf{First Maria multiplies equation(1) by 2:}\n2(2x+4y)=2(10)\n\textsf{or, }4x+8y=20........(3)\n\n\textsf{Then she adds the resulting equation to equation (3):}\n(4x+8y)+(3x-y)=20+8\n\textsf{or, }7x+7y=28

\textsf{Maria is wrong because she obtained the wrong equation.}

Two angles that can be trisected with a straightedge and compass are a 90angle and a _____ angle.A. 9
B. 75
C. 45
D. 90

Answers

The correct answer is:

C) 45.

Explanation:

An angle of measure (2\pi)/(N) can be trisected if and only if N is not divisible by 3.

A 45° angle in radians is (\pi)/(4). Setting up a proportion, we can find the value of N to determine if this fits:

(\pi)/(4)=(2\pi)/(N)

Cross multiply:

\pi * N = 4 * 2\pi \n N\pi = 8\pi
 
This means N = 8, and is not divisible by 3; thus the 45 degree angle is trisectible.

the answer on apex is 45°

Matilda is filling a circular section of her porch with colorful pebbles. She knows the diameter of the circular section is 15ft, and each bag of pebbles covers 10ft. How many bags of pebbles will Matilda need? A. 15 bags B. 16 bags C. 17 bags D. 18 bags

Answers

Answer:

D. 18 bags

Step-by-step explanation:

Given:

Matilda is filling a circular section of her porch with colorful pebbles.

Diameter of circular section = 15 feet.

Each bag of pebbles covers 10 square feet.

Question asked:

How many bags of pebbles will Matilda need?

Solution:

Diameter of circular section = 15 feet

Radius, r = (Diameter)/(2)=(15)/(2) =7.5\ feet

Area\ of\ circular\ section=\pi r^(2)

                                        =(22)/(7) *7.5*7.5=(1237.5)/(7)=176.78} \ square\ feet

10 square feet is covered by number of bag of pebbles = 1

1 square feet is covered by number of bag of pebbles = (1)/(10)

176.78 square feet is covered by number of bags of pebbles =  

(1)/(10)*176.78=17.67

We found up to 17 bags of pebbles there is still some place left to fill the section:-

Thus, Matilda need about 18 bags of pebbles to fill the  circular section of her porch.

Lim[(sqrt(cos(2x))-cubicrootof(cos(x))]/arctg(x^2) when x->0

Answers

\lim\limits_(x\to0)\frac{√(cos2x)-\sqrt[3]{cosx}}{arctanx^2}=\left[(0)/(0)\right]\n\nDe\ L'Hospital's\ rule:\n\n\lim\limits_(x\to0)\frac{√(cos2x)-\sqrt[3]{cosx}}{arctanx^2}=\lim\limits_(x\to0)\frac{(√(cos2x)-\sqrt[3]{cosx})'}{(arctanx^2)'}=(*)\n-------------------------------\n(√(cos2x)-\sqrt[3]{cosx})'=(-sin2x)/(√(cos2x))-\frac{-sinx}{3\sqrt[3]{cos^2x}}=\frac{-2sinxcosx\cdot3\sqrt[3]{cos^2x}+sinx√(cos2x)}{3√(cos2x)\cdot\sqrt[3]{cos^2x}}\n\n(arctanx^2)'=(2x)/(x^2+1)

-------------------------------\n(*)=\lim\limits_(x\to0)\frac{-6sinxcosx\sqrt[3]{cos^2x}+sinx√(cos2x)}{3\sqrt[3]{cos^2x}\cdot√(cos2x)}:(2x)/(x^4+1)\n\n=\lim\limits_(x\to0)\frac{-sinx(6cosx\sqrt[3]{cos^2x}-√(cos2x))}{3\sqrt[3]{cos^2x}\cdot√(cos2x)}\cdot(x^4+1)/(2x)\n\n=\lim\limits_(x\to0)(-sinx)/(x)\cdot\frac{(6cosx\sqrt[3]{cos^2x}-√(cos2x))(x^4+1)}{6}=-1\cdot\frac{(6\cdot1\sqrt[3]{1^2}-\sqrt1)(0+1)}{6}\n\n=-1\cdot(5)/(6)=-(5)/(6)

A round cake pan that is 2 inches deep and has a 7 inch diameter a 6 inch x9 inch rectangular cake pan that is 2 inches deep a. which of these pans has the larger volume justify your answerb. excluding the bottom of the cake what is the surface area of each cake that will be covered by frosting justify your answer

Answers

a) Pan 2 has a larger volume because:

Pan 1:

Circle area = pi x r^2
7/2 = 3.5
3.5 x 3.5 = 12.25inch^2
12.25 x pi = 38.48451
38.48451 x 2 = 76.96902inch^3 (76.97inch^3 to 2dp)

Pan 2:

6 x 9 = 54inch^2
54 x 2 = 108inch^3

b)

Pan 1:

Surface area covered =  60.48inch^2 (2dp) because:

Circle area = pi x r^2
7/2 = 3.5
3.5 x 3.5 = 12.25inch^2
12.25 x pi = 38.48451inch^2

Circumference = pi x diameter
pi x 7 = 21.9911486
21.9911486inch^2

21.9911486 + 38.48451 = 60.4756586inch^2

Pan 2:

Surface area covered = 114inch^2 because:

Bottom:
6 x 9 = 54inch^2

2 longer sides:
2 x 9 = 18
18 x 2 = 36inch^2

2 shorter sides:
2 x 6 = 12
12 x 2 = 24inch^2

Total = 54 + 36 + 24 = 114inch^2

(Hope this helps.)
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