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mixer [17]
2 years ago
7

You have a function that takes in an X value and produces a Y value. The y value equals 24 times the x value, plus 4 more. What'

s the y value when the x value equals 4?
Mathematics
1 answer:
lapo4ka [179]2 years ago
6 0

Answer:

100

Step-by-step explanation:

First, write out the equation

y = 24x + 4

Then plug in x = 4 to solve for y

y = 24x + 4

y = (24)(4) + 4

y = 100

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You are paying all eight of your monthly recurring bills free online now that you used to pay by "snail-mail". Stamps cost 47 ce
Ede4ka [16]

Answer:

  $45.12

Step-by-step explanation:

  ($0.47 /stamp)×(8 stamps/mo)×(12 mo/yr) = $45.12 /yr

You will save $45.12 per year on stamps.

8 0
2 years ago
Carolina is mowing lawns for a summer job. For every mowing job, she charges an initial fee plus \$6$6dollar sign, 6 for each ho
Margarita [4]

For this case, the first thing we must do is define variables.

We have then:

t: number of hours

F (t): total charge

We write the function that models the problem:

F (t) = 6t + b

Where,

b: represents an initial fee.

We must find the value of b.

For this, we use the following data:

Her total fee for a 4-hour job, for instance, is $ 32.

We have then:

32 = 6 (4) + b

From here, we clear the value of b:

32 = 24 + b

32-24 = b

b = 8

Then, the function that models the problem is:

F (t) = 6t + 8

Answer:

the function's formula is:

F (t) = 6t + 8

7 0
2 years ago
Evaluate variable expressions involving integers
murzikaleks [220]
What are the expressions
3 0
2 years ago
A daffodil grows 0.05m every day. Plot the growth of the flower if the initial length of the daffodil is 0.8m and hence give the
Vanyuwa [196]

Answer:

1.2m

Step-by-step explanation:

You must first find out how much the daffodil grew over the 8 days:

0.05 x 8 = 0.4

Then you must add how much it grew to the original height:

0.4 + 0.8 = 1.2

Hope this helps you out! : )

8 0
2 years ago
A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto
Gre4nikov [31]
\begin{cases}y=2x-2\\y=2x+2\end{cases}\implies\begin{cases}-2x+y=-2\\-2x+y=2\end{cases}

For these lines, let u=-2x+y.

\begin{cases}y=2-x\\y=4-x\end{cases}\implies\begin{cases}x+y=2\\x+y=4\end{cases}

And for these, let v=x+y.

Now,

\begin{cases}u=-2x+y\\v=x+y\end{cases}\implies \begin{bmatrix}u\\v\end{bmatrix}=\underbrace{\begin{bmatrix}-2&1\\1&1\end{bmatrix}}_{\mathbf T}\begin{bmatrix}x\\y\end{bmatrix}

The vertices of S in the x-y plane are (0, 2), (2/3, 10/3), (2, 2), and (4/3, 2/3). Applying \mathbf T to each of these yields, respectively, (2, 2), (2, 4), (-2, 4), and (-2, 2), which are the vertices of a rectangle whose sides are parallel to the u-v plane.
6 0
2 years ago
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