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IRISSAK [1]
4 years ago
6

Fred Has A bag Of Sweets:

Mathematics
1 answer:
Grace [21]4 years ago
4 0
A: 1/20 (number of black sweet divided by total number of sweets)
b: Green (convert 1/4 to common denom of 20 (5/20), therefor sweets which equal 5 are green.
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Show your work and check answer.<br> -3x=5x+8
frutty [35]
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4 years ago
Read 2 more answers
Which is the graph of f(x) = log3x?.
valentinak56 [21]

Logarithm functions are the inverses of the exponential function the graph of the given function is attached in the given image.

The points for the given function are (1,0) (3,1) and (9,2)

Given that

Function; \rm  f(x) = log3x

We have to determine

The graph of the function f(x).

According to the question

The logarithmic function is the inverse of the exponential function.

Function; \rm  f(x) = log_3x

To plot the graph of the given function we have to find the asymptotes;

Vertical asymptote at x =0

The point at x= 1 is,

\rm f(1) = log_3(1)\\\\f(1)=log_33\\\\f(1)=1

The point at x= 3 is,

\rm f(3) = log_3(3)\\\\f(3)=log_33\\\\f(3)=1

The point at x= 9 is,

\rm f(9) = log_3(9)\\\\f(9)=3log_33\\\\f(9)=3\times 1\\\\ f(9)=3

Hence, the points for the given function are (1,0) (3,1) and (9,2).

To know more about the Exponential function given below.

brainly.com/question/3653847

5 0
3 years ago
Find the sine pls help
Nat2105 [25]

Answer:

7/25 is correct answer dude

7 0
3 years ago
A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t
Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square  at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

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V(x)=x*(26-x)*(13-x)

To get maximum volume differentiate the above equation,

V(x)=x*(26*13-26*x-13*x+x^2)

V(x)=x^3-39x^2+338x\\

V'(x)=3x^2-78x+338

V''(x)=6x-78

Now ,Solve the Quadratic Equation to get x values,

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x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

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x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17  or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0  i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

V(x)=x^3-39x^2+338x

V(x)=3.8288^3-39*(3.82883)^2+338*3.8288

V(x)=56.13-571.73+1294.13

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3 years ago
A rectangle has a width of 2cm and a perimeter of 25cm. Find the length and the area .
Gwar [14]

2w + 2l = 25

2 * 2 + 2l = 25

<em><u>Simplfiy.</u></em>

4 + 2l = 25

<em><u>Subtract 4 from both sides.</u></em>

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l = 10.5

Now that we have the value of l, we can find the area.

A = lw

A = 10.5 * 2

A = 21.

The length is 10.5cm.

The area of the rectangle is 21cm^2

5 0
3 years ago
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