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maria [59]
3 years ago
9

meyer used 6 loads of gravel to cover 2/5 of his driveway.how many loads of gravel will he need to cover his entire driveway?

Mathematics
2 answers:
Virty [35]3 years ago
8 0
Meyer will need 15 loads of gravel to cover his entire driveway.
natka813 [3]3 years ago
6 0
He will need 13, and will have extra
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3.) Graph f(x)=3^x.​
ollegr [7]

Answer:

f(x)=3^x plot = see attachment

Step-by-step explanation:

7 0
3 years ago
How do you sketch a graph of a function?
Illusion [34]

The graph of a function can be sketched by plotting points.

First and foremost, it is to be determined if the function is odd/even or periodic. Next, we find the x and y intercepts. To sketch a graph of a function, one can start by plotting a few points that lie on the graph using the coordinates given by the function. For example, if the function is y = x², then points (0, 0), (1, 1), and (2, 4) can be plotted by substituting the values of x into the function to find the corresponding values of y.

Subsequent to this, one can connect the points with a smooth curve to form the graph of function. If the function is a straight line, the graph will be a straight line. If the function is quadratic, the graph will be a parabola. The graph may take on a more complicated shape for more complex functions.

Read more about the graph of a function on:

brainly.com/question/26857518

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4 0
1 year ago
HELP ME!!!! ASAP!!!!! Thank you so much!!!!
likoan [24]
3/2x + 1/5 >= -1
3/2x >= - 6/5
x >= -12/15
x >= -4/5

-1/2x - 7/3 >= 5

-1/2x >= 22/3
x <= -44/3

Its D
5 0
3 years ago
Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

8 0
3 years ago
Solve the system of equations.<br> 2.5y + 3x = 27<br> 5x – 2.5y = 5
worty [1.4K]
X=4 and y=6
For this, you should use simultaneous equations
3 0
3 years ago
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