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3 years ago

Y = 5.3x, when x = 10

Mathematics

1 answer:

Sveta_85 [38]3 years ago

4 0

the answer is 53 because when you multiply a decimal by 10 all you need to do is move the decimal to the right

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If F(x)= x2+2x-3 and g(x)=x2-9 , find (f/g)(4) and(f+g)(4) .(f/g)(4) is , and (f+g)(4) is .

Alexandra [31]

F(x) = x2+2x-3 = 1
g(x) = x2-9 = -7

(f/g)(4) = (1/-7)(4) = (-0.14)(4) = -0.57
(f+g)(4) = (1+(-7))(4) = (-6)(4) = -24

7 0

4 years ago

Can y’all plz help me ?

crimeas [40]

Answer:

x=-5

Step-by-step explanation

The first box has eight x's and six 1's which means that it would be written as 8x+6

The second box has four x's and fourteen -1's which means that it would be written as 4x-14

You set them equal to one another

8x+6=4x-14

You subtract 4x from both sides

4x+6=-14

Subtract 6 on both sides

4x=-20

Divide 4 on both sides

x=-5

Hope it helps :))

4 0

3 years ago

Pjpjpjpjpjpj<br>pjpjpjpjpjpj

bogdanovich [222]

Answer:

Hmm tough question I'm pretty sure the answer is pj.

Step-by-step explanation:

In the question they are very upfront with the answer, it's a trick question to make you believe it's hard or there is no answer. So, with that being said the answer is pj.. Happy to help anytime. :)

8 0

3 years ago

Read 2 more answers

What matrix results from −8·A?

djverab [1.8K]

Is that all the question says can you give it into more details

So I can help you out

3 0

3 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago