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3 years ago

6

The green team wants to beautify the schools garden. As part of the beautification process, the green team is building a fence a

round the garden. What is the total amount of fence they will need to complete the project?

1 answer:

Pepsi [2]3 years ago

6 0

the answer is 10 yeeet

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Find the measure of each side indicated. Round to the nearest tenth.

choli [55]

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I'm say d last choice is yours answer

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3 years ago

Please help! algebra 1 math! whoever gets right is Brainliest! (10 points)

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3 years ago

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Marie wants to know the most popular car for parents. She plans to survey the people entering the mall from 10:00 a.m. to 11:00

Andreyy89

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C. Marie should make sure she surveys both the fathers and mothers.

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2 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

Which of the following pairs of functions are inverses of each other?

Serggg [28]

Answer:

<em>The answer is (B).</em>

Step-by-step explanation:

A). y = f(x) = 6(x - 2) + 3 ⇔ y = 6x - 12 + 3 ⇔ 6x = y + 9 ⇔ x = $\frac{y+9}{6}$

g(x) = $\frac{x+9}{6}$ is inverse function of f(x) and ≠ $\frac{x+2}{6}$ - 3

<em>B).</em> y = f(x) = $\frac{5x}{4}$ - 3 ⇔ 4y = 5x - 12 ⇔ x = $\frac{4y+12}{5}$

g(x) = $\frac{4(x + 3)}{5}$ is inverse of f(x)

6 0

3 years ago