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3 years ago

In a candy mix there are 4 green bars for every 3 red bars.how many red bars are there if there are 200 green bars

Mathematics

1 answer:

Viktor [21]3 years ago

7 0

150 red bars, since 200 divided by 4 equals 50, then 3 x 50 would give u 150 red bars

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I need help with these two questions please !

Lapatulllka [165]

For the first question the answer is B. multiply 357.43 by 12 you'll get the answer $4,289.

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3 years ago

Read 2 more answers

What is 3/8x + 15/2=18

mart [117]

Let's solve for x --

Isn't it always a little easier to solve a math problem with starting by rewriting the equation?

3/8x + 15/2 = 8

Step 1) Subtract 15/2 from both sides

3/8x + 15/2 - 15/2 = 8 - 15/2

3/8x = 1/2

Step 2) Multiply both sides by 8/3

(8/3) × (3/8x) = (8/3) × (1/2)

x = 4/3

FINAL ANSWER -

x = 4/3

↑ ↑ ↑ Hope this helps! :D

3 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

Plzzzzzzz helllp failing

telo118 [61]

Answer:

i belive a or c please tell me if i waz wroung

7 0

3 years ago

Please I really need help on this one

Salsk061 [2.6K]

Answer:

2.5ft

Step-by-step explanation:

d=2r

r=1/2d

r=1/2*5ft=2.5 ft

6 0

3 years ago