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3 years ago

9

charlene has eight more dollars than Megan. Together they have $86.00. Which equiation can be used to determine the amount of mo

ney, m, Megan has?

1 answer:

Paul [167]3 years ago

8 0

Answer:

Megan has 35.00 dollars.

Step-by-step explanation:

Divide 86.00 by 2 which equals 43.00. Next, minus 8 from the 43.00 and you will get what Megan has. Megan has 35.00 dollars.

I hope this helps!

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In this situation, choose length and width such that 12.75=l*w. writing it as a mixed number you have 51/4=sqrt51/2*sqrt51/2 or any other combination you can use.

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3 years ago

A ride-share company has a fee that includes a fixed cost and a cost that depends on both the time spent travelling, in minutes,

zlopas [31]

Answer:

the cost of Roy's ride is $23.05

Step-by-step explanation:

According to the Question,

Let, Cost of per minute charge is 'x' & Cost Of Per Kilometre charge is y .

Given, A ride-share company has a fee of the fixed cost of a ride is $2.55 .
And, The Total cost of the Ride depends on both the time spent on travelling(in minutes), and the distance travelled(in kilometres) .

⇒ Judy's ride costs $16.75 . but the actual cost after deducting the fixed charge is 16.75-2.55 = $14.20, took 8 minutes & The distance travelled was 10 km. Thus, the equation for the journey is 8x+10y=14.20 ⇒ Equ. 1

⇒ Pat's ride costs $30.35 . but the actual cost after deducting the fixed charge is 30.35-2.55 = $27.80, took 20 minutes & The distance travelled was 18 km. Thus, the equation for the journey is 20x+18y=27.80 ⇒ Equ. 2

Now, on Solving Equation 1 & 2, We get

x=0.4(Cost of per minute charge) & y=1.1(Cost Of Per Kilometre charge)

Now, Roy's ride took 10 minutes & The distance travelled was 15 km . Thus, the cost of Roy's Ride is 10x+15y ⇔ 10×0.4 + 15×1.1 ⇔ $20.5

Hence, the total cost of Roy's ride is 20.5 + 2.55(fixed cost) = $23.05

8 0

3 years ago

Graph the image of this figure after a dilation with a scale factor of 1/3 centered at the origin.

posledela

Answer:

see attached diagram

Step-by-step explanation:

A dilation with a scale factor of $\dfrac{1}{3}$ centered at the origin has a rule:

$(x,y)\mapsto\left(\dfrac{x}{3},\dfrac{y}{3}\right).$

Then the image of the triangle ABC is triangle DEF(see attached diagram) with coordinates

$D\left(\dfrac{-3}{3},\dfrac{3}{3}\right)\Rightarrow D(-1,1);$
$E\left(\dfrac{3}{3},\dfrac{9}{3}\right)\Rightarrow E(1,3);$
$F\left(\dfrac{6}{3},\dfrac{6}{3}\right)\Rightarrow F(2,2).$

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3 years ago

Sean has a collection of coins. One tenth of the coins are from Europe. Thirty-two hundredth are from Asia. The rest are from Af

Tema [17]

Lets write each part, the total will be x:
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3200 = Asia
x - <span>(1/10)x - 3200 = Africa

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= (320x + 1)/3200
that would be expressed as a fraction, and depends on the total of coins, x</span>

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3 years ago

Read 2 more answers

How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t

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By first principles, the derivative is

$\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h$

Use the binomial theorem to expand the numerator:

$(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n$

$(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

The first term is eliminated, and the limit is

$\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h$

A power of $h$ in every term of the numerator cancels with $h$ in the denominator:

$\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)$

Finally, each term containing $h$ approaches 0 as $h\to0$ , and the derivative is

$y=x^n\implies y'=nx^{n-1}$

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3 years ago