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2 years ago

What’s every 4 out of 8

Mathematics

1 answer:

klio [65]2 years ago

7 0

Answer:I didn't know exactly what you meant sooo... here are a few ways.

<h2 /><h2>If it is a fraction, 1/2</h2><h2>if it is a decimal, .50 </h2><h2>if it is ratio, 1:2</h2><h2>If it is subtraction, 4</h2>

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James has eight coins in his pocket. All the coins are either dimes or quarters.The total value of the coins is $1.25. How many

Goshia [24]

Answer:

he has 5d+3q

Step-by-step explanation:

since every d is .10 and every q is .25 the equation can be changed to 5(.10)+3(.25) which is equal to .5+.75 which adds up to 1.25

i did this by assuming that the amount of d is 8 at first, then change one of the 8 d to a q which will give you 7d+q, then assume another d is a q which will give you 6d+2q, repeat this until the amount of d and q add up to 1.25

6 0

3 years ago

Can anyone help ? :(

V125BC [204]

Answer: B 65 and C 275

Step-by-step explanation:

Just divide by 3 then multiply by 5

7 0

3 years ago

How to solve x/-3-2=5.5

amid [387]

Okay first set up the equation

x
--- -2 = 5.5
-3

add over the 2

x
--- = 7.5
-3

now you need to multiply the denominator by the answer

(-3)(7.5) = x
-22.5 = x

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

can you help me evaluate this: 7m + 14m - 6n - 5n + 2m. why do you add 6n with 5n instead of subtract?

Mademuasel [1]

Step-by-step explanation:

The reason is because it is not just 6 it is - 6 so when there are two negative numbers they don't subtract they add if it was 6n then it was to be subtracted

21m - 11n + 2m

23m - 11n

6 0

3 years ago