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2 years ago

Do all the dimes or all the nickels have a greater total value

1 answer:

3 0

That depends on how many of each you have.

-- If the number of dimes is more than half the number of nickels,
then the dimes have more value.

-- If the number of nickels is more than double the number of dimes,
then the nickels have more value.

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The Roman numeral which represents the number in standard form for 3,089 is?

Tresset [83]

MMMLXXXIX

~ hope this helps!

4 0

2 years ago

PLS HELP<br> Solve the equation for a. Show your work. <br> 4a-14=7a-8

OLEGan [10]

A=-2 Im attaching a picture with the steps

7 0

2 years ago

Read 2 more answers

HELP NEEDED ASAP! Thanks :)

Arlecino [84]

Answer:

Step-by-step explanation:

First we must find the total boxes of pens

15 + 9 = 24

Then we can divide the total cost by the total number of boxes of pens

$72 / 24 boxes = 3

4 0

2 years ago

I need answer on this graphing va substitution.Problem 7 I tried to solve it but not sure that’s correct

Bond [772]

Given the system of equations:

$\begin{gathered} 2x+9y=27 \\ x-3y=-24 \end{gathered}$

To solve it by substitution, follow the steps below.

Step 1: Solve one linear equation for x in terms of y.

Let's choose the second equation. To solve it for x, add 3y to each side of the equations.

$\begin{gathered} x-3y=-24 \\ x-3y+3y=-24+3y \\ x=-24+3y \end{gathered}$

Step 2: Substitute the expression found for x in the first equation.

$\begin{gathered} 2x+9y=27 \\ 2\cdot(-24+3y)+9y=27 \\ -48+6y+9y=27 \\ -48+15y=27 \end{gathered}$

Step 3: Isolate y in the equation found in step 2.

To do it, first, add 48 to both sides.

$\begin{gathered} -48+15y=27 \\ -48+15y+48=27+48 \\ 15y=75 \end{gathered}$

Then, divide both sides by 15.

$\begin{gathered} \frac{15y}{15}=\frac{75}{15} \\ y=5 \end{gathered}$

Step 4: Substitute y by 5 in the relation found in step 1 to find x.

$\begin{gathered} x=-24+3y \\ x=-24+3\cdot5 \\ x=-24+15 \\ x=-9 \end{gathered}$

Answer:

x = -9

y = 5

or (-9, 5)

Also, you can graph the lines by choosing two points from each equation, according to the picture below.

7 0

11 months ago

Evaluate using L'Hopital's rule:<br><br> <img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%201%5E%7B%2B%7D%7D%20%28x%20-%201

SSSSS [86.1K]

Answer:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1$

Step-by-step explanation:

We are given:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)$

And we want to evaluate it using L'Hopital's Rule.

First, using direct substitution, we will acquire:

$=(1-1)^\ln(1)}=0^0$

Which is indeterminate.

In order to apply L'Hopital's Rule, we first need to manipulate the expression. We will let:

$y=(x-1)^\ln(x)$

By taking the natural log of both sides:

$\ln(y)=\ln(x)\ln(x-1)$

And by taking the limit as x approaches 1 from the right of both functions:

$\displaystyle \lim_{x\to 1^+}\ln(y)=\lim_{x\to 1^+}\ln(x)\ln(x-1)$

Rewrite:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{\ln(x-1)}{\ln(x)^{-1}}$

Using direct substitution on the right will result in 0/0. Hence, we can now apply L'Hopital's Rule:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}\Big(\frac{-x\ln(x)^2}{-x\ln(x)^2}\Big)$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-\frac{x\ln(x)^2}{x-1}$

Now, by using direct substitution, we will acquire:

$\displaystyle \Rightarrow -\frac{1\ln(1)^2}{1-1}=\frac{0}{0}$

Hence, we will apply L'Hopital's Rule once more. Utilize the product rule:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-(\ln(x)^2+2x\ln(x))$

Finally, direct substitution yields:

$\Rightarrow -(\ln(1)^2+2(1)\ln(1))=-(0+0)=0$

Thus:

$\displaystyle \lim_{x\to 1^+}\ln(y)=0$

By the Composite Function Property for limits:

$\displaystyle \lim_{x\to 1^+}\ln(y)=\ln( \lim_{x\to 1^+}y)=0$

Raising both sides to e produces:

$\displaystyle e^{\ln \lim_{x\to 1^+}y}=e^0$

Therefore:

$\displaystyle \lim_{x\to 1^+}y=1$

Substitution:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1$

4 0

2 years ago