3 years ago

It takes 7 weeks to build 8.75 kilometers. At what rate did they build it?

Mathematics

1 answer:

Brrunno [24]3 years ago

8 0

$8.75 \div 7 = 1.25$
They built 1.25km every week

You might be interested in

I(1, 1) and J (-3,-3)<br> Find the midpoint of the line segment

Pavlova-9 [17]

The Answer is (-1,-1)

4 0

3 years ago

What is the result when 75,069 is divided by 45

Lera25 [3.4K]

Answer: 1668.2

Multiply 1668.2×45 and get 75,069

5 0

3 years ago

Read 2 more answers

A stone is dropped from the top of a tower. The height h in meters from where the stone is dropped to the ground is given by the

CaHeK987 [17]

Wag umasa sa brainly

Step-by-step explanation:

Labot

5 0

2 years ago

Today Cenisa runs 450 meters more than yesterday. Yesterday she ran 1.23 kilometers. How many meters does Cenisa run today?

aalyn [17]

Answer: it’s 1680 meters, i did this question today.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago