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lapo4ka [179]
3 years ago
7

Which is the graph of f(x) = 5(2)Y?

Mathematics
2 answers:
Setler79 [48]3 years ago
6 0

Answer:

4

Step-by-step explanation:

ale4655 [162]3 years ago
5 0

Answer:

It does not have a graph.

Step-by-step explanation:

I think that the question was asked incorrectly.

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The sum of three numbers is 104. The first number is 10 less than the second. The third number is 4 times the second. What are t
Shalnov [3]

Answer:

The numbers are 9, 19 and 76.

Step-by-step explanation:

Let's take the second number and make it a variable, like x. Now let's find all the other numbers in terms of x.

The first number in terms of x:

x - 10

The second number in terms of x:

4x

If we add them all up, these 3 numbers in terms of x must be equal to 104, so:

x + (x - 10) + (4x) = 104 \\ x + x - 10 + 4x = 104 \\ 6x - 10 = 104 \\ 6x = 114 \\ x = 19

Now that we know the second number, we can find the other numbers since we already figured out their values in terms of x:

19 - 10 = 9

4 \times 19 = 76

The first number is 9, second number 19 and the third number 76.

6 0
3 years ago
A student ran 100 m dash in 15.4 seconds. What was the student speed in miles per hour
amid [387]

15.4 seconds = 0.00428 hours and 100 meters = 0.062 miles soo the student ran approximately 14.5 mph

8 0
3 years ago
Solve for x..........10 points*
Ivanshal [37]

Answer:

\huge\boxed{\sf x = 12}

Step-by-step explanation:

8x + 34 = 130 <u>(Vertical Angles are Equal)</u>

Subtract 34 to both sides

8x = 130 - 34

8x = 96

Divide both sides by 8

x = 96 / 8

x = 12

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
3 0
3 years ago
Read 2 more answers
State the extremes of the proportion below. 4/7=12/x
yKpoI14uk [10]

\frac{4}{7}  =  \frac{12}{x}  \\  \\ 1. \:  \frac{4}{7}x  = 12 \\  \\ 2. \:  \frac{4x}{7}  = 12 \\  \\ 3. \: 4x = 12 \times 7 \\  \\ 4. \: 4x = 84 \\  \\ 5. \: x =  \frac{84}{4}  \\  \\ 6. \: x = 21

7 0
3 years ago
Read 2 more answers
The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

  • A(0) = 10.
  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

4 0
3 years ago
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