$21,000 at 3% for 3 years

Karo-lina-s [1.5K]

Sorry if these are not correct, but this is what i came up with

1. -36 is less
2. Decimal is 0.12 and fraction is 3/25
3. About 3 1/3 quartz of paint (not too sure)
4. Commutative property of addition
5. Associative Property of multiplication
6. Additive identity
7. Distributive property
8. Multiplicative Property of Zero

4 0

2 years ago

Read 2 more answers

Explain Each Part please!!!

ch4aika [34]

Constant of Variation has this formula for Direct Variation.
y = kx
Where y = cost in dollars; x = length in meters; k = constant

We will use the formula for direct variation because the problem states that the cost varies directly to the length.

In the problem we need to get the constant of variation.

k = y/x
k = $4500 / 200 meter
k = $22.50 cost per meter.

A.) The equation relating C and s is:

k = C/s
* i just substituted C for y and s for x.

B.) The constant of variation represents the additional cost that the company will pay for every meter in length they add. which is $22.50 per meter.

C.) y = kx ; substitute C and s to y and x respectively;

C = ks
C = $22.50 * 700 m
C = $15,750

8 0

3 years ago

Read 2 more answers

3. A wire of length 90 cm is cut into three pieces. The lengths of the pieces are in the ratio 3:5:7. Calculate the length of (a

Nata [24]

the difference between the longest and the shortest piece

4 0

2 years ago

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Evaluate using L'Hopital's rule: <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

Is (-1, -1) a solution to this system of equations? y = -8x - 9 y = -5x – 6 yes no

ollegr [7]

Answer:

Yes

Step-by-step explanation:

To see is (-1, -1) is a solution to the system, we can plug it into the system to check:

-1 = -8(-1)-9

-1 = 8-9

-1 = -1

-1 = -5(-1)-6

-1 = 5-6

-1 = -1

Both of the equations are true, therefore (-1, -1) is a solution to the system.

8 0

2 years ago

$21,000 at 3% for 3 years​

$21,000 at 3% for 3 years