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3 years ago

What number is three tenths less than 1?

Mathematics

2 answers:

miss Akunina [59]3 years ago

8 0

Answer:

7/10 or seven/tenths

liraira [26]3 years ago

7 0

7 tenths or seven tenths because if you count down in the tenths and one is a whole number 5...

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Y varies directly with x. If x = 16 and y = 2 find y when x = 96

ArbitrLikvidat [17]

Answer:

y = 12

Step-by-step explanation:

Use the equation y = kx

Plug in x and y to find k:

2 = k(16)

1/8 = k

Then, plug in 1/8 as k and 96 as x to find y:

y = 1/8(96)

y = 12

6 0

3 years ago

Help please And how would you get the answer need asap

Rashid [163]

No the 2 equations are not equal
the -1/10, -1/10 can't be subtracted from 1 3/10g
they are not like terms
the proper equation is
1/5g converts to 2/10g
2/10g - 1g + 1 3/10 g = 5/10g or 1/2g
1/2g - 2/10 is final equation

4 0

3 years ago

Morgan borrowed $0.35 from her mother every time she went to the gas station to buy some chips. Over the summer Morgan went to t

GarryVolchara [31]

Answer:

The money borrowed by Morgan is $ 3.15.

Step-by-step explanation:

Money borrowed by Morgan to buy chips = $ 0.35

Number of times goes to the station = 9

Money borrowed by Morgan in 9 time visit to station

= 9 x $ 0.35

=$ 3.15

5 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

Okay so this is super tricky, there is a riddle on my math quiz. Yk for extra credit and its 25% of my grade, according to Mr. W

nlexa [21]

If there was just one of each animal, the outcome would be that of 1 cow and 1 dog.

Option(B) is correct.

The average cost, or how much it generally costs to produce a unit, can be discovered using the cost function.

If you own two cats worth $100 each and a cow worth $2,500. One dog, though, is worth $7000.

Cost of 1 cow = $2500

Cost of 2 cats = $100

Thus, cost of 1 cat = $100/2 = $50

Cost of 1 dog = $7000

If there was only 1 of each animal, total cost = $(2500+50+7000) = $9550.

This outcome resembles the situation of having 1 cow and 1 dog.

Learn more about cost here:

brainly.com/question/25109150

#SPJ1

3 0

2 years ago