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faltersainse [42]
3 years ago
12

Someone please help me with #6 .

Mathematics
1 answer:
weeeeeb [17]3 years ago
6 0
The answer is c because you plug in the values given. 4 times .75 is three. Three plus three is 6. For the other side you plug in .50. 6 times .50 is 3. 3-2 is 1. Then you are asked to find the perimeter, a rectangle has 4 sides so you do 6 on the top and bottom and 1 on each of the sides. When you add those all together you get 14 which is c.

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ABCD is a rectangle.
grandymaker [24]

Answer:

i need more details

Step-by-step explanation:

8 0
3 years ago
A right triangle whose hypotenuse is 3 centimeters long is revolved about one of its legs to generate a right circular cone. Fin
SSSSS [86.1K]

Answer:

The height of the right circular cone when constructed this way is \sqrt3 cm.

The radius of the right circular cone when constructed this way is \sqrt6 cm.

The volume of the right circular cone when constructed this way is 6\sqrt3 \pi cm³.

Step-by-step explanation:

Given that,

A right triangle whose hypotenuse is 3 cm long is revolved .

Then other two legs of the triangle will be radius and height of the cone.

Assume the height and radius of the cone be h and r respectively.

From Pythagorean Theorem :

h²+r²=3²

⇒ r²= 9 - h²

Then the volume of the cone is

V= π r²h

⇒ V= π(9-h²)h                [ ∵ r²= 9 - h²]

⇒V= π(9h - h³)

Differentiating with respect to h

V'=π(9 - 3h²)

Again differentiating with respect to h

V''= π(-6h)

⇒V''= (-6πh)

To maximum or minimum ,we set V'=0

π(9 - 3h²)=0

⇒3h²=9

⇒h²=3

\Rightarrow h=\sqrt3

Now, V''|_{h=\sqrt3}=-6\pi (\sqrt3).

Since at h=\sqrt3,V''<0.

The volume of cone is maximum at h=\sqrt3 cm when constructed this way.

The height of the right circular cone when constructed this way is \sqrt3 cm.

The radius of the right circular cone when constructed this way  r=\sqrt{9-(\sqrt3)^2

   =  \sqrt{9-3}

   =\sqrt6 cm.

The volume of the  right circular cone when constructed this way is

=π r²h

=\pi (\sqrt6)^2\sqrt3

=6\sqrt3 \pi cm³

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=2y%20%3D%20xm%20%2B%20%28%20%5Csqrt%7Bw%20%2B%202gm%7D%20%29%20%5Cdiv%20%7Bw%7D%5E%7B2%7D%20"
KiRa [710]

Answer:

m = -(2g - 4xy)/2x²w⁴ + √((2g - 4xy)² - 4*(x²w⁴)*(4y²w⁴ - w))/2x²w⁴

Step-by-step explanation:

2y= xm + √(w + 2gm) ÷ w^2

multiply all through by w²

2yw² = xm(w²) + √(w + 2gm)

rearranging

√(w + 2gm) = 2yw² - xmw²

square both sides

w + 2gm = (2yw² - xmw²)²

w + 2gm = (2yw² - xmw²)*(2yw² - xmw²)

w + 2gm = 4y²w⁴ - 2yxmw⁴ - 2yxmw⁴ + x²m²w⁴

collecting like terms

(x²w⁴)m² + (2g - 4yxw⁴)m + (4y²w⁴ - w). = 0

This seems to have been transformed to a quadratic equation

so we solve the quadratic equation by formula taking m as the variable

For simplicity, put

x²w⁴ = a

2g - 4xy = b

4y²w⁴ - w = c

we have

am² + bm + c = 0

solving the quadratic by formula

(-b + √(b² - 4ac))/2a. (taking the positive part)

-b/2a +√(b² - 4ac)/2a = m

substituting back

x²w⁴ = a

2g - 4xy = b

4y²w⁴ - w = c

m = -(2g - 4xy)/2x²w⁴ + √((2g - 4xy)² - 4*(x²w⁴)*(4y²w⁴ - w))/2x²w⁴

6 0
4 years ago
Is the correct answer to this problem one solution because 19x=-19?? Or am I wrong??
nadezda [96]
I think you are right so far what i've done comes out to your answer.
6 0
3 years ago
This pattern repeats every 5 terms. The pattern is trapezoid, square, triangle, pentagon, square. What is the 17th term of this
qaws [65]

Answer:

b Square

Step-by-step explanation:

I literally just counted

3 0
3 years ago
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