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3 years ago

I need help plz ...i need this done by today !!!!

Mathematics

1 answer:

Deffense [45]3 years ago

6 0

5. At First Street Elementary School, about 18% of the 610 students ride bicycles to school. About how many students ride bicycles to school?

Since 100% is equal to 610 students

Then 18% is equal to X students

Results X students = 18%*610students/100% = 109.8 students. As it does not make sense to split students, we have to round up to the nearest unit. Then we have 110 students. That would be approximately 18.03%, that rounded to the nearest unit is approximately 18%.

Answer 5) 110 students ride bicycles to school.

6. Write a proportion and answer each question using the conversion factor 1 inch = 2.54 centimeters.

a) Don is 67 inches tall. How many centimeters tall is he?

Since 1 inch is equal to 2.54 centimeters

Then 67 inches will be X centimeters

Results X cm = 67in * 2.54cm/1in = 170.18 centimeters

Answer 6a) Don is 170.18 centimeters tall.

b) Amira's pencil is 13.5 centimeters long. How many inches long is it?

Since 2.54cm is equal to 1inch

Then 13.5cm will be X inches

Results X inches = 13.5cm * 1in/2.54cm ≈ 5.315 approximately 5.32 inches

Answer 6b) Amira's pencil is approximately 5.32 inches long.

7. Use dimensional analysis to convert 60 miles per hour to feet per second. Remember that there are 5,280 feet in 1 mile.

Since 1 mile is equal to 5,280 feet

Then 60 miles will be X feet

Results X feet = 60 miles * 5,280 feet/1 mile = 316,800 feet

Since there are 3,600 seconds in 1 hour

316,800 feets per hour = 316,800 feets/hour x 1hour/3,600secs = 88 feet/sec

Answer 7) 60 miles per hour are converted to 88 feets per second

Hope this helps!

$\textit{\textbf{Spymore}}$

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points A,F, & G

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They lie on same line l. By definition they are collinear.

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3 years ago

Read 2 more answers

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Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
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Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

3 years ago

Mother dairy sold 3,355 litres of milk in the month of April and May if they sold equal amount every day what was its daily sale

Shkiper50 [21]

Answer:

<h2>55 litres </h2>

Step-by-step explanation:

Litres of milk sold by the dairy in April and May together = 3,355

So,

Litres of milk sold by the dairy per day = ?

To find this we have to do addition and division and have to go through the calender a bit :)

<h2>Let's solve it!!</h2>

No. of days in April = 30 days

No. of days in May = 31 days

Therefore total no. of days in both the months = 30 + 31

= 61

Now,

Litres of milk sold by the dairy per day ( or in simple words the daily sale )

$= \frac{3355}{61}$

$= 55$

Si therefore 55 litres of milk was the daily sale.

Hope it helps you!!

#IndianMurga(. ❛ ᴗ ❛.)

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