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Rus_ich [418]
2 years ago
9

7.

Mathematics
1 answer:
ch4aika [34]2 years ago
8 0
The answer is option d via completing the square

y= x^2+2x+1

y= (x+1)^2+1

so it will have an aos of -1 and a vertex of (-1,0)
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(-1/4)^8 * (8/11)^6 * (1/2)^-8
marissa [1.9K]

Answer:

1024/1771561

Step-by-step explanation:

Using exponent rules, we can multiply all of them together after applying the exponent to get the final answer.

8 0
3 years ago
Find the perimeter of triangle PQR
Katyanochek1 [597]
27a+5.9+12.2 =Perimeter
5 0
2 years ago
Describe the graph of the function f(x) = x3 − 11x2 + 36x − 36. Include the y-intercept, x-intercepts, and the shape of the grap
kap26 [50]
1) y-intercept => x = 0, => y = f(0) = 0 - 0 + 0 - 36 = -36

2) x-intercept => y = 0 => factor the function (start by dividing by x -2)

f(x) = (x-2)(x-3)(x-6) =0 => x =2, x = 3, x = 6 (these are the x-intercepts)

3) critical points:

between x = 2 and x = 3, there is a local maximum

between x =3 and x = 6 there is a local minimum

3) Shape.

The function comes growing from - infinity.

In the third quadrant the function is negative (it does not pass throuhg the second quadrant)

It enters to the fourth quadrant intercepting the y-axis at y = -36. It continues growing and intercepts the x-axis at x = 2.

 It continues increasing until a maximum local positive value, starts to decrease, intercepts the x-axis at x = 3, continues decreasing, becomes negative, gets a local minimum in the fourth quadrant, starts to increase, intercepts the x-axis at x = 6, becomes positive, and continues growing.
3 0
3 years ago
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
3. A children's slide is 10 feet long and inclines 64° from the ground. The
Nata [24]

Answer:

  87.0°

Step-by-step explanation:

The law of sines can be used to solve this. We have two sides of a triangle and the angle opposite one of them. We want to find the angle opposite the other known side.

In the attached, the triangle is ΔACS. We have side "a" = 9, and side "c" = 10. Angle A is given as 64°. The law of sines tells us ...

  sin(C)/c = sin(A)/a

  sin(C) = (c/a)sin(A)

  C = arcsin((c/a)sin(A)) = arcsin(10/9·sin(64°)) ≈ 87.03°

The ladder makes an angle of about 87° with the ground.

3 0
2 years ago
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