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Studentka2010 [4]
3 years ago
15

Find the average rate of change for f(t)=6t2+7 over the interval [4,4+h]

Mathematics
1 answer:
Eddi Din [679]3 years ago
6 0
Average rate of change using formula
\frac{f(4 + h) - f(4)}{(4 + h) - (4)}
Just plug in and simplify!
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X solve for the following inequality-1/2 p less than -16 which graph shows the correct solution
valkas [14]

Answer:

See attachment

Step-by-step explanation:

The given inequality is

-  \frac{1}{2}p \: <  \:  - 16

We multiply both sides by -2 and reverse the inequality sign to get:

- 2 \times  -  \frac{1}{2} p \:   >  \:  - 2 \:  \times  - 16

p \:  >  \: 32

We draw an open circle at 32 and draw an arrow to the right as shown in attachment.

8 0
3 years ago
Read 2 more answers
Which transformations can be used to map a triangle with vertices A(2, 2), B(4, 1), C(4, 5) to A’(–2, –2), B’(–1, –4), C’(–5, –4
Romashka [77]
The triangles ABC and A'B'C' are shown in the diagram below. The transformation is a reflection in the line y=-x. This is proved by the fact that the distance between each corner ABC to the mirror line equals to the distance between the mirror line to A'B'C'.

6 0
3 years ago
What is 250/12 plus 100/21 add all that up to get your answer
Dafna1 [17]

Answer:

  1075/42

Step-by-step explanation:

  \dfrac{250}{12}+\dfrac{100}{21}=\dfrac{250(21)+(12)100}{(12)(21)}=\dfrac{6450}{252}\\\\=\dfrac{(6)(1075)}{(6)(42)}=\boxed{\dfrac{1075}{42}}

8 0
3 years ago
Value for -3 + 4x &gt; 4
lilavasa [31]

Answer:

x > 1/4

Step-by-step explanation:

6 0
3 years ago
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
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