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3 years ago

02119a rectangular pool is 7 feet wide. it is 3 Times as long as it is wide

Mathematics

1 answer:

marshall27 [118]3 years ago

5 0

What are you looking for?? Area? Perimeter?

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24 students in a class took an algebra test If 18 students passed what percent passed

Anuta_ua [19.1K]

Answer: 75%

We know that 1/4 of 24 (25%) is 6, 6 fits into 18 three times and 25 times 3 is 75. So the answer is 75%

6 0

2 years ago

2x + 3y =10 x + y = 3 please do workings asap

kompoz [17]

This is system of Equations.
2x + 3y =10
x + y = 3

x + y = 3 (multiply this by three) to get 3x + 3y = 9
2x + 3y =10
3x + 3y = 9 Subtract them

-1x = 1
x=-1
Now plug x back into the top equation and solve for y.

I hope this helps! If not, let me know! Have a fantastical day!

5 0

3 years ago

Read 2 more answers

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

Fred is making a bouquet of carnations and roses. The carnations cost $5.25 in all. The roses cost $1.68 each. How many roses di

olga nikolaevna [1]

Answer:

8 roses

Step-by-step explanation:

First start by subtracting $18.69 by $5.25 because you already know how much the carnations cost and how much the bouquet cost in all. When you subtract 18.69 by 5.25 you should get $13.44. If each rose cost $1.68 then divide 13.44 by 1.68 to get the amount of roses in the bouquet.

8 0

3 years ago

Volume of prisms mathswatch answers

Svetradugi [14.3K]

Answer:

Volume of recangular prisms is l*w*h, triangular prism is (l*w*h)/3

Step-by-step explanation:

8 0

3 years ago