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2 years ago

Karen wants to string lights around the edge of her deck. The shape and dimensions of her deck are shown in the diagram

Mathematics

1 answer:

Marianna [84]2 years ago

8 0

She would need 70 feet of lights

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which represents the solution(s) of the system of equations, y=x^2-2x-15 and y=8x-40? determine the solution set algebraically

Harrizon [31]

y=x^2-2x-15 (1) y=8x-40 (2)

8x-40=x^2-2x-15
X^2+10x-25=0
(x-5)^2
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3 years ago

Curtis decided to go on a road trip to Canada. On the first day of his trip, he drove for 15 hours and traveled 1,005 miles. At

Radda [10]

Answer:

Step-by-step explanation:

1,005 / 15 = 67

3 0

3 years ago

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You are on a ski trip with your friends. Your friend Meg asks you to lend her $250. Write down the conditions that will help you

Arisa [49]

whether you have the money to lend her

whether she offers you some collateral

whether she is creditworthy

whether she can repay the loan within a period that suits you

whether she offers to pay interest

4 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

Please I really need help! I will mark you as brainliest!

iogann1982 [59]

Answer:

-10 / -5 / -1 / 0 / 1 / 5 / 10

Step-by-step explanation:

I hope this helps!

5 0

3 years ago