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aksik [14]
2 years ago
14

At 1:00 P.M., the the water level is 28 inches. the water level in a pool is 13 inches. At 1:30 P.M., the water level is 18 inch

es. At 2:30 P.M., the water level is 28 inches. What is the constant rate of change?
Mathematics
2 answers:
Artemon [7]2 years ago
6 0

Answer:

10 inches per hour

Step-by-step explanation:

The water level was 13 inches at 1 PM. In 30 minutes (So in 1:30 PM) the water level was 18 inches, so it had increased by 5 inches in 30 minutes (Which is 10 inches in 1 hour). The rate of change is how much it has risen per time, we can also check that between 1;30 pm and 2:30 pm, which are 1 hour apart, the water level had risen from 18 inches to 28 inches, which is 10 inches in an hour.

Dmitrij [34]2 years ago
5 0

Answer:

the answer is 19.667

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Which shows how to solve the equation Three-fourths x = negative 6 for x in one step?
My name is Ann [436]

option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).

<u>Step-by-step explanation:</u>

We have , The given expression as Three-fourths x = negative 6  , which can be written as \frac{3}{4} (x) = -6 . Now in order to solve this equation in one step , we must notice that coefficient of x must be 1 but it's  \frac{3}{4} , Let's make coefficient of x  as 1 by multiplying both side of equations by 4/3:

\frac{3}{4} (x) = -6

⇒ \frac{3}{4} (x) = -6

⇒ \frac{3}{4}(\frac{4}{3} ) (x) = -6(\frac{4}{3} )

⇒ x = -6(\frac{4}{3} )

⇒ x = -2(4)

⇒ x = -8

Therefore, x= -8 & correct option to solve the equation Three-fourths x = negative 6 for x in one step is <u>option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).</u>

7 0
2 years ago
Read 2 more answers
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
Arjay, Dorothy, Melissa, and Gray live in the same city. Arjay and Dorothy live 2 miles from each other. Dorothy and Melissa liv
GarryVolchara [31]
Dorothy and Gray live 8 miles from each other.
5 0
2 years ago
2 Points
hichkok12 [17]

Answer: 345.02

Step-by-step explanation:

8 0
2 years ago
An elevator in a tall building goes up 7 floors, then
3241004551 [841]

Answer: he started on 14th floor

Step-by-step explanation:

14 + 7 - 9 - 4 + 8 - 2= 14

It started on floor 14

5 0
2 years ago
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