Give one example of when you would use the friendly numbers strategy to subtract

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

7 0

When there are one positive and one negative

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The cost of 5cd's and a $12 DVD

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Let the price of one CD be c.
The 5 CDs cost 5c.
The cost of 5 CDs and a $12 DVD is

5c + 12

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3 years ago

Henry has x blue pens. Rachel has $5$ blue pens and $y$ black pens. Jake has $3$ times the total number of pens Henry and Rachel

Troyanec [42]

Answer:

Step-by-step explanation:

Henry X blue

Rachel 5 blue and Y black

Jack 3 times the total of pens ( blue & black)

X + 5 + Y for Henry and Rachel

Jack : 3( X + 5 + Y)

Jack has 3X + 15 + 3Y.

Pens of jack 3X + 15 Blue & 3Y black.

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2 years ago

(1 point) a tank contains 1060 l of pure water. a solution that contains 0.06 kg of sugar per liter enters the tank at the rate

kirill115 [55]

(a) There is 0 kg of sugar in the tank at the beginning since it contains pure water at the start. The sugar only comes from the solution.

(b)

$S' = f(t,S) = \left(0.06 \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) - \left(\dfrac{S}{1060} \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) \ \Rightarrow \\ \\ S' = 0.54 \text{ kg}/\text{min} - \dfrac{9S}{1060}$

So yes, you enter S' = 0.54 - (9S/1060)

(c)

$\displaystyle\frac{dS}{dt} = 0.54 - \frac{9S}{1060} \ \Rightarrow\ \frac{dS}{dt} = \frac{572.4 - 9S}{1060}\ \Rightarrow\ \dfrac{dS}{572.4 - 9S} = \frac{1}{1060} dt\ \Rightarrow \\ \\
\int \dfrac{dS}{572.4 - 9S} = \int \frac{1}{1060} dt\ \Rightarrow\textstyle\ -\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t + C \\ \\
S(0) = 0 \ \Rightarrow\ -\frac{1}{9}\ln|572.4 - 0| = \frac{1}{1060}(0) + C\ \Rightarrow\ C = -\frac{1}{9} \ln 572.4$

$-\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t -\frac{1}{9} \ln 572.4\ \Rightarrow \\ \\
\ln|572.4 - 9S| = \ln 572.4 - \frac{9}{1060}t \ \Rightarrow \\ \\
|572.4 - 9S| = e^{\ln 572.4 - 9t/1060}\ \Rightarrow \\ \\
572.4 - 9S= \pm 572.4 e^{-9t/1060}\ \Rightarrow \\ \\
S = \frac{-1}{9}\left(-572.4 \pm 572.4 e^{-9t/1060}\right)$

But only (+) satisfies $S(0) = 0$

$S= -\frac{1}{9}\left(-572.4 + 572.4 e^{-9t/1060}\right) \\ \\
S= 63.6 - 63.6 e^{-9t/1060}\text{ kg}$

Enter in S = 63.6 - 63.6 * e^(-9t/1060)

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I know how to translate an object but this is kinda tricky.

solniwko [45]

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so
A' (5, -3)
then reflected across y-axis
A' (-5, -3)

a)A(4,2), point A is located in quadrant I
b)A'(5,-3), point A' is located in quadrant IV
c)A'(-5,-3), point A' is located in quadrant III

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(2y^4)^3 fffffffffffffffffffffffffff

qwelly [4]

Best answer is 8y^12

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