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3 years ago

Ok so these math questions are simple. Would you be able to do them?

Mathematics

1 answer:

alexandr402 [8]3 years ago

3 0

Answer:

3) it would be the first one 3*(6/5) and it would be the second one 3/(5*6)

Step-by-step explanation:

when you put it in the calculator they come out to be 3.6

I have no idea for the question above it

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Tim has a savings account with the bank. The bank pays him 1% per year. he has $5,700 and wondered when it will reach $5,800. Wh

Reika [66]

Answer:

Step-by-step explanation:

$A=P(1+rt)\\ \\ t=\left(\frac{A}{P}-1\right)/r\\ \\ t=\left(\frac{5800}{5700}-1\right)/0.01=1.75y$

6 0

3 years ago

Danika is making necklaces.she has 512 silver beads and 278 blue.About how many more silver than beads does Danika have?

exis [7]

Daniela has 234 more silver beads than blue beads

5 0

3 years ago

Read 2 more answers

What is not equivalent to 36/72

STALIN [3.7K]

Any fraction that does not equal 1/2.

6 0

3 years ago

Can any one help me lease I do not under stand.

svetoff [14.1K]

Answer: the cube root of 512 is 8.

Step-by-step explanation:

First divide 512 and take the least factors as I have shown,

512= 2×2×2×2×2×2×2×2×2

Now, make the triple twins of these factors,

(2×2×2)(2×2×2)(2×2×2)

Now, take one number from each pair and multiply them,

2×2×2=8

Therefore, the answer of the cube root of 512 is 8.

4 0

3 years ago

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

Ok so these math questions are simple. Would you be able to do them?​

Ok so these math questions are simple. Would you be able to do them?