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3 years ago

11

What numbers are divisible by 3

2 answers:

steposvetlana [31]3 years ago

7 0

Tbh idrk but uh thanks for the points

creativ13 [48]3 years ago

7 0

Answer:

An easy trick to find out: A number is divisible by 3 if the sum of its digits is divisible by 3. For example, 375, 3+7+5=15. 15 is divisible by 3, so 375 is too.

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1 1/3 x 2 3/4 = <br> ANSWER ASAP AND BE RIGHT <br> EXPLAINNNN STEP BY STEP <br> TO BE BRAINLIEST

sladkih [1.3K]

Answer:

1/2

Step-by-step explanation:

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3 years ago

Express the radical using the imaginary unit, i

Kipish [7]

Answer:

7i.

Step-by-step explanation:

Note : The given expression must be $\sqrt{-49}$ .

We have to write this radical value in the imaginary value.

The given value can be rewritten as

$\sqrt{-49}=\sqrt{49\times (-1)}$

$\sqrt{-49}=\sqrt{49}\times \sqrt{-1}$ $[\because \sqrt{ab}=\sqrt{a}\sqrt{b}]$

$\sqrt{-49}=7\times i$ $[\because \sqrt{-1}=i]$

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3 years ago

What is the original price of ATV set if you paid 1035 for it after a 10% discount

oksano4ka [1.4K]

Answer:

$1150

Step-by-step explanation:

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5 0

3 years ago

Which expression could be used to determine the cost of a $40 video game after a 15 percent discount?

LUCKY_DIMON [66]

Answer: The required expression is,

$ 40(0.85)

Step-by-step explanation:

Given,

The original cost of the videogame = $ 40,

Discount percentage = 15%,

So, the amount of discount = 15% of 40

= (0.15)40 ( ∵ 1% = 0.01 )

Hence, the cost of the videogame = original cost - discount

= 40 - (0.15)40

= 40 (1-0.15)

= 40(0.85)

Which is the required expression .

5 0

4 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

2 years ago