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2 years ago

Robert wants to purchase a skateboard for $109.98. He will also have to pay a 6.5% sales tax,

Mathematics

1 answer:

Furkat [3]2 years ago

6 0

Answer:

$117.13

Step-by-step explanation:

Hope this answers your question and have a great day!

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If you have a question or a comment USE THE QUESTION COMMENTS GOD DANG IT

kow [346]

Answer:

8/9

Step-by-step explanation:

In the standard form of line y = mx + b

m is the slope and b is the y-intercept

so just compare these two equations and you will see that the slope here is just 8/9

4 0

2 years ago

Ellie missed 76 questions on her multiple choice history final and earned a grade of 62%. How many total questions were on the f

erica [24]

Answer:

200 questions on the final exam.

Step-by-step explanation:

38% = 76 questions, 100% = 2.63157894737 x 76 = 200

(100/38=2.63157894737 hence this is the answer)

6 0

1 year ago

You are buying a $50 video game as a holiday gift for your brother luckily you have not just one but two coupons for that store

postnew [5]

Whats the question though?

6 0

3 years ago

Which expression is equivalent to 15a^8b^4/5a^4b

LekaFEV [45]

Answer: The given expression $\frac{15a^8b^4}{5a^4b}$ simplified to $3a^4b^3$

Step-by-step explanation:

Given : expression $\frac{15a^8b^4}{5a^4b}$

We have to simplify the given expression $\frac{15a^8b^4}{5a^4b}$

Consider the given expression $\frac{15a^8b^4}{5a^4b}$

Divide the numbers $\frac{15}{5}=3$

$=\frac{3a^8b^4}{a^4b}$

Apply exponent rule, $\frac{x^a}{x^b}\:=\:x^{a-b}$

We have,

$\frac{a^8}{a^4}=a^{8-4}=a^4$

$=\frac{3a^4b^4}{b}$

Cancel out common factor b,

We have

$=3a^4b^3$

Thus, the given expression $\frac{15a^8b^4}{5a^4b}$ simplified to $3a^4b^3$

4 0

3 years ago

Read 2 more answers

A population of insects increases at the rate of 200 10t 13t2 what is the change in the population of insects between day 0 and

RUDIKE [14]

Answer:

762 days

Step-by-step explanation:

Given

$Rate = 200 + 10t + 13t^2$

Let the rate be R.

So the rate change with time is represented as:

$\frac{dR}{dt} = 200 +10t + 13t^2$

So:

$dR = (200 + 10t + 13t^2)\ dt$

To get the number of insects between day 0 and day 3, we need to integrate dR and set the bounds to 0 and 3

i.e.

$dR = (200 + 10t + 13t^2)\ dt$ becomes

$\int\limits^3_0 {dR} \, =\int\limits^3_0 (200 + 10t + 13t^2)\ dt$

$R =\int\limits^3_0 (200 + 10t + 13t^2)\ dt$

Integrate

$R = 200t + \frac{10t^2}{2} + \frac{13t^3}{3} [3,0]$

Solve for R by substituting 0 and 3 for t

$R = (200*3 + \frac{10*3^2}{2} + \frac{13*3^3}{3}) - ( 200*0 + \frac{10*0^2}{2} + \frac{13*0^3}{3})$

$R = (200*3 + \frac{10*9}{2} + \frac{13*27}{3}) - (0 + \frac{10}{2} + \frac{0}{3})$

$R = (200*3 + 5*9 + 13*9) - (0)$

$R = 600 + 45 + 117 - 0$

$R =762$

<em>The population of insect between the required interval is 762</em>

4 0

3 years ago