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2 years ago

Martin needs 60 meters of fence to surround a rectangular garden.

Mathematics

1 answer:

Karolina [17]2 years ago

6 0

Answer:

The garden is 10 ft. wide.

Step-by-step explanation:

2 • (2w + w) = 60 here is an equation to show how I got the width (10 ft.)

2 x 3w = 60

6w = 60

Divide 6 from both sides, 6 & 6 cancel out and 60 divided by 6 is 10.

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ANTONII [103]

Answer:

19 orders of tea

57 orders of collee

Step-by-step explanation:

find 1/4 of the 76

76/4=19

multiply by 3

19x3=57

8 0

3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

Step 2: Find Right-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

Step 3: Find Left-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ Since $\displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)$ , then $\displaystyle \lim_{x \to 5} f(x) = DNE$

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

Unit: Limits

5 0

3 years ago

Tim and Tom are painters. Tim and Tom are painting a crate. Tom paints 10 sq ft per minute. They painted a particular crate in 1

slava [35]

Tom-------------- >10 sq ft per minute

Tim------------------- >3.5*10=35 sq ft per minute

Both painters (10+35)=45 sq ft per minute

let's assume a cubic unit crate----------> V1=1*1*1=1 feet3

the total surface of the cube is S1=6*(1*1)=6 sq ft

if the volume of the cube increases twice

V2=2 feet3 ------------------- > 2=x*x*x------------- >x=2^(1/3)

And the S2=6*(2^(1/3)* 2^(1/3)=9.5244 sq ft

V2/V1=9.5244/6=1.5874

Therefore

If Tim and Tom----------------------- > take 1 day----------- > to paint a crate V1

For the crate V2------------- > take 1.5874 days

assuming an 8-hour work day

0.5874*8=4.70 hr-------------- > 4 h 42 min

The answer is 1 day 4h 42 min

6 0

4 years ago

Name the edges of an hexagonal pyramid

aleksandrvk [35]

Given:

Hexagonal pyramid

To find:

The edges of an hexagonal pyramid.

Solution:

Edges means lines which connecting to vertices.

Edges in the base of the pyramid:

AB, BC, CD, DE, EF, FA

Edges in the triangular shape of the pyramid:

AG, BG, CG, DG, EG, FG

Therefore edges of an hexagonal pyramid are:

AB, BC, CD, DE, EF, FA, AG, BG, CG, DG, EG, FG

4 0

4 years ago