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3 years ago

7

A store in Roger's neighborhood sells boxes of pencils that have 3 pencils in each box .Roger bought several boxes of pencils at

the store . Which could be the number of pencils he bought.

2 answers:

musickatia [10]3 years ago

5 0

He could have had 3,6,9,12,15,,18,21,24,27,30 etc....

Vladimir [108]3 years ago

4 0

Write down on a piece of paper 3,6,9,12,15,18,21,24,27,30 and for sure u'll find the answer

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An electronics store usually sells computers priced at $1500 each. If the customer orders the computer over the Internet, he has

iragen [17]

Answer:

There would be a 20% price decrease, making the price $300 less.

7 0

3 years ago

Read 2 more answers

Find g'(t) for the function g(t) = 7/t^4

Alex17521 [72]

The differentiation of the function g(t) = 7/t⁴ will be equal to g¹(t)=-28/t⁵

<h3>What is differentiation?</h3>

The method of determining the derivative, or rate of change, of a function in mathematics.is termed as the differentiation.

Given that:-

$g(t)= \dfrac{7}{t^4}$

The derivative will be calculated as:-

$g'(t)=7 \dfrac{d}{dt}(t^{-4})\\\\\\g'(t)=7\times (-4t^{-5})\\\\\\g'(t)=\dfrac{-28}{t^5}$

Therefore the differentiation of the function g(t) = 7/t⁴ will be equal to g¹(t)=-28/t⁵

To know more about derivatives follow

brainly.com/question/954654

#SPJ1

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2 years ago

Suppose that the function f ( x ) = 2x + 12 represents the cost to rent x movies amonth from an internet movie club. Makayla now

Dafna11 [192]

$x=7\ \ \ \ \Rightarrow\ \ \ f(x)=2\cdot7+12=14+12=26\ [\$]\\\\\$26-\$10= \$16\\\\Ans.\ Makayla\ need\ \ \ \$16.$

3 0

3 years ago

Read 2 more answers

Need help with this. Hope someone can help

Triss [41]

Hi,

Let me help you with this problem.

$v = ( \frac{2 \times 1.5}{2} ) \times 3 \\ v = \frac{3}{2} \times 3 \\ v = 1.5 \times 3 \\ v = 4.5$

Answer: 4.5m3

Hope this helps.
r3t40

7 0

4 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago