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Bad White [126]

3 years ago

6

NIne times ten to the negative power of three

1 answer:

Ivanshal [37]3 years ago

3 0

9 x 10⁻³ is the same amount as 0.009

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What’s the area of 36 square meters perimeter of 30 meters what’s the width and length of those?

Anvisha [2.4K]

≠ means not equal to
L x w = 36
2L + 2w = 30

1. Find the multiples of 36

1 and 36
2 and 18
3 and 12
4 and 9
6 and 6

2. Figure out which set of numbers multiplied by 2 equals 30.

For 1 and 36:
2(1) + 2(36) = 2 + 72 = 74 ≠ 30

For 2 and 18:
2(2) + 2(18) = 4 + 32 = 36 <span>≠ 30

For 3 and 12:
2(3) + 2(12) = 6 + 24 = 30

So your answer is 3 and 12</span>

3 0

3 years ago

Bill is making accessories for the soccer team. He uses 838.95 inches of fabric on headbands for 32 players and 3 coaches. He al

creativ13 [48]

There was 23.29 inches used on a headband for each player.

And 8.54 inches used on a wristband for each player.

To find out how much fabric for headbands and would beused for each player/person you would do

So, if you substitute the values in it is

And finally, to find how much fabric is used on a wristband for each player/person you would use the same formula.

7 0

2 years ago

in a fundraising committee of 90 people the ratio of men to women is 7:2. find the number of women required to join the existing

LuckyWell [14K]

Hello,

Step-by-step explanation:

90 = 9 x 10 = (7 + 2) x 10 = 70 + 20

70 men and 20 women

$\frac{70}{5} = \frac{20 + X}{4}$

56 = 20 + X ⇔ X = 36

Answer : + 36 women

7 0

3 years ago

Round 2,539 to the nearest tenth

daser333 [38]

Answer:2,540

Step-by-step explanation:

6 0

3 years ago

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