All categories

2 years ago

Isabella filled her pool with water at a constant rate.

Mathematics

2 answers:

AlladinOne [14]2 years ago

6 0

(184-94)/5=18

This means that she fills up 18 liters every minute. Thus she will be done in->

184(original amount)+(2 minutes (time passed during first data point)*18(rate of water)=184+36=220

To find the total time taken we divide 220 by 18 which is 220/18=12 2/9

eimsori [14]2 years ago

3 0

<h2>Answer:</h2>

The pool is filling at the rate of 18 liters per minute.

Step-by-step explanation:

We have been given time in minutes and Volume in liters. We can find the speed at which the volume changes by finding the slope.

Formula for slope will be :

$\frac{v2-v1}{t2-t1}$

Choosing any 2 values from table. (7,94) and (12,4) denoted as (t1, v1) and (t2, v2)

Calculating the slope we get,

$\frac{4-94}{12-7}$

= $\frac{-90}{5}=-18$

This means the pool is filling at the rate of 18 liters per minute.

You might be interested in

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

KIM [24]

Answer:

The answer is

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$tan(\theta)=-1$

Step-by-step explanation:

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$sin^{2}(\theta)+cos^{2}(\theta)=1$

In this problem we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

$\frac{3\pi}{2}$

The angle $\theta$ belong to the third or fourth quadrant

The value of $sin(\theta)$ is negative

Step 1

Find the value of $sin(\theta)$

Remember

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(\theta)=1-\frac{1}{2}$

$sin^{2}(\theta)=\frac{1}{2}$

$sin(\theta)=-\frac{\sqrt{2}}{2}$ ------> remember that the value is negative

Step 2

Find the value of $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$tan(\theta)=-1$

8 0

2 years ago

Read 2 more answers

A pet store owner, Byron, needs to determine how much food he needs to feed the animals. Byron knows that he needs to order the

Brums [2.3K]

Answer:

He needs to buy 120 packages of dod food.

Step-by-step explanation:

b - bird food

h - hamster food

d - dog food

c - cat food

b + h + d + c = 600

b = h (bird food is the same as hamster food)

b = 4 d (bird food is 4 times dod food)

c = d/2 (cat food is half dog food)

Total of packages ordered.

b + b + d + c = 600

4d + 4d + d + d/2 = 600

4 d + 4 d + d + d = 600. 2

10 d = 1.200

d= 120 .........the amount of dog food packages.

6 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

2 years ago

.580 80 repeating as fraction

denis-greek [22]

First off, we'll move the non-repeating part in the decimal to the left-side, by doing a division by a power of 10.

then we'll equate the value to some variable, and move the repeating part over to the left as well.

anyhow, the idea being, we can just use that variable, say "x" for the repeating bit, let's proceed,

$\bf 0.580\overline{80}\implies \boxed{\cfrac{5.80\overline{80}}{10}}\qquad \textit{now, let's say }x= 5.80\overline{80}\\\\
-------------------------------$

$\bf thus\qquad \begin{array}{llll}
100\cdot x&=&580.80\overline{80}\\
&&575+5.80\overline{80}\\
&&575+x
\end{array}\qquad \implies 100x=575+x
\\\\\\
99x=575\implies x=\cfrac{575}{99}\qquad therefore\qquad \boxed{\cfrac{5.80\overline{80}}{10}}\implies \cfrac{\quad \frac{575}{99}\quad }{10}
\\\\\\
\cfrac{\quad \frac{575}{99}\quad }{\frac{10}{1}}\implies \cfrac{575}{99}\cdot \cfrac{1}{10}\implies \cfrac{575}{990}\implies \stackrel{simplified}{\cfrac{115}{198}}$

and you can check that in your calculator.

4 0

2 years ago

Read 2 more answers

Katy had two choices of routes to get her to work. She wanted to choose the route that would get her to work fastest, on average

zmey [24]

Answer:

D. There is not enough evidence at the 5% significance level to indicate that one route gets Katy to work faster, on average, since 0 falls within the bounds of the confidence interval.

Step-by-step explanation:

At 5% confidence level, Katy found difference in mean commuting times (Route 1-Route 2) in minutes as (-1,9).

Since no difference in means (0 min) falls within the confidence level (-1,9), we can not reject the hypothesis that there is no difference in mean commuting times when using Route1 or Route2.

A higher significance level(10% etc) may lead a shorter confidence interval leaving 0 outside and may reach a conclusion that Route1 takes longer than Route2

4 0

2 years ago