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3 years ago

Simplify the following expression completely using distribution and combining like terms

Mathematics

1 answer:

LiRa [457]3 years ago

5 0

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

Simplify ⇨ 19m + 14 - 3 (3m + 7)

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\sf \: 19 m - 14 - 3 ( 3 m + 7 )$

Use the distributive property to multiply -3 by 3m+7.

$\sf \: 19m-14-9m-21$

Combine 19m and -9m to get 10m.

$\sf \: 10m-14-21$

Subtract 21 from -14 to get -35.

$\boxed{\boxed{\bf \: 10m-35 }}$

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Divide 20 chocolates between Adi and Suresh such that adi gets 60% while Suresh gets 40% of the chocolates

Eddi Din [679]

Hello!

To find how much chocolate Adi and Suresh receives, you need to convert the percentage to a decimal, then you multiply that decimal by the total number of chocolates, which is 20.

To convert percentages to decimals, divide the percentage by 100, and remove the percentage.

60% / 100 = 0.60

20% / 100 = 0.20

Then, we multiply 0.60 and 0.20 by 20.

0.60 x 20 = 12

0.40 x 20 = 8

Therefore, Adi gets 12 chocolates and Suresh gets 8 chocolates.

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3 years ago

10. Mike bought five items for lunch that cost a total of $4.55. Hot dogs cost $1.10 each

Reil [10]

Answer:

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Step-by-step explanation:

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3 years ago

If you have 3/4 cups of flour how many eighths do you have

Shtirlitz [24]

Just do 3/4 divided by 1/8 - answer should be 6

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3 years ago

Suppose the linear regression line y= 4.009x- 77.531 predicts a pizza parlor's profits based on the number of pizzas sold. If x

tamaranim1 [39]

Answer:

The estimated profit is 62.784

Step-by-step explanation:

y = 4.009x - 77.531

Plug in the number of pizzas sold for x:

y = 4.009(35) - 77.531 = 140.315 - 77.531 = 62.784

5 0

3 years ago

Read 2 more answers

A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and

allsm [11]

(a) Let $A(t)$ denote the amount of sugar in the tank at time $t$ . The tank starts with only pure water, so $\boxed{A(0)=0}$ .

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min

so that the net rate of change of $A(t)$ is governed by the ODE,

$\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}$

$A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}$

Multiply both sides by the integrating factor $e^{7t/1080}$ to condense the left side into the derivative of a product:

$e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}$

$\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}$

Integrate both sides:

$e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt$

$e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C$

Solve for $A(t)$ :

$A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}$

Given that $A(0)=0$ , we find

$0=\dfrac{378}5+C\implies C=-\dfrac{378}5$

so that the amount of sugar at any time $t$ is

$\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}$

(c) As $t\to\infty$ , the exponential term converges to 0 and we're left with

$\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5$

or 75.6 kg of sugar.

7 0

3 years ago