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2 years ago

Help me please I don’t know this stuff

Mathematics

1 answer:

yuradex [85]2 years ago

7 0

Answer: 24 because jeff earned $600 and for jason to earn the same amount 24*25 hours = 600

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What is the algebraic expression for "the difference between eleven times a number and four"?

marshall27 [118]

11n-4

I believe that is correct

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3 years ago

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How do you write 2.51 in expanded form

wel

2.0 + 0.5 + 0.01 hope this helps

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2 years ago

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Three friends Kofi,Yaw and Musa are in partnership business in which Kofi and Yaw contributed 5/9 and 2/15 of the cost of 14000.

wlad13 [49]

Answer:

(a) $Total = 45000$

(b) Yaw, Musa and Kofi

Step-by-step explanation:

Given

$Kofi =5/9$

$Yaw = 2/15$

$Musa = 14000$

Solving (a): The cost of the business

First, we solve for the fraction of Musa

$Musa + Kofi + Yaw = 1$

$Musa = 1 - (Kofi + Yaw )$

$Musa = 1 - (5/9 + 2/15)$

$Musa = 1 - (\frac{25+6}{45})$

$Musa = 1 - \frac{31}{45}$

$Musa = \frac{45 - 31}{45}$

$Musa = \frac{14}{45}$

So, we have:

$\frac{14}{45} * Total = 14000$

Make Total the subject

$Total = 14000 * \frac{45}{14}$

$Total = 1000 * 45$

$Total = 45000$

Solving (b): Partners in ascending order

First, represent the fractions as decimals

$Kofi =5/9$ $= 0.5556$

$Yaw = 2/15$ $= 0.1333$

$Musa = \frac{14}{45}$ $= 0.3111$

From the conversion above, the least is 0.1333 (Yaw), then 0.3111 (Musa), then 0.5556 (Kofi).

<em>So, the order is: Yaw, Musa and Kofi</em>

6 0

3 years ago

HELP QUICK PLEASE!! WILL GIVE BRAINLIEST ANSWER

Alex

F(12)=-34(12)+2
f(12)=-408+2
f(12)=-406

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3 years ago

Calculate the total area of the shaded region.

LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$

7 0

2 years ago