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3 years ago

У= 4 +X у у Х 2 3 4 5 Х

Mathematics

2 answers:

mr Goodwill [35]3 years ago

7 0

What would you like me to help with?

Allisa [31]3 years ago

3 0

Answer:

what am i supposed to do?

Step-by-step explanation:

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The Jelly Junior High school color is made by mixing red paint with yellow paint. The ratio of red to yellow is 3 to 5. How much

user100 [1]

V/a = 3/5
a = 20
v = ...

3/5 = v/20
5v = 20*3
5v = 60
v = 60/5
v = 12 oz

8 0

4 years ago

Which math expression means "44 more than an unknown number"?

worty [1.4K]

Answer: C. X + 44

Step-by-step explanation:

Let x be the unknown number.

Given statement : "44 more than an unknown number".

By using addition operation ,

Mathematical expression for "44 more than an unknown number" = 44 +x

Hence, the correct expression is x+44.

So, the correct option is C.

5 0

3 years ago

If 7 is the mean of 10,3,4,a,6,9 and 10.find the value of a

nydimaria [60]

Answer:

a = 7

Step-by-step explanation:

I think, what you need to do, is first understand how to find the mean of anything: It's quite simple, actually, just add up all the numbers, then divide that number by the number of numbers. So, let's put it in perspective:

10, 3, 4, a, 6, 9, 10

7 is the mean. 7 is the number of numbers.

If, to find the mean is: 10 + 3 + 4 + a + 6 + 9 + 10= 7a ÷ 7, then: 10 +3 + 4+ 6 + 9 +10= 42

42 plus what divided by 7 is the question now.

42 + a ÷ 7= 7

So, 7 times 7, that's 49 minus 42 and you've got your answer: 7

8 0

3 years ago

Is 9/16 closest to 0, 1/2, 1, or 9

REY [17]

Answer:

1/2

Step-by-step explanation:

9/16 is equal to 0.5625 . The closest number to this is 0.5 which is the same as 1/2

8 0

3 years ago

Read 2 more answers

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

3 years ago