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3 years ago

Find the volume of this square pyramid

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

5 0

Answer:

216

Step-by-step explanation:

Svet_ta [14]3 years ago

4 0

Answer:

72yd

Step-by-step explanation:

Hope that helps

hsobsnsjns

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Write in summation notation.

Ivahew [28]

If that is -2+1+4+13 I think it should be 16

3 0

3 years ago

Determine the value of c so that f(x) is continuous on the entire real line when

mixer [17]

Answer:

A. -4

Step-by-step explanation:

Given the function f(x) = x + 3 for x ≤ -1 and 2x - c for x > -1, for the function to be continuous, the right hand limit of the function must be equal to its left hand limit.

For the left hand limit;

The function at the left hand occurs at x<-1

f-(x) = x+3

f-(-1) = -1+3

f-(-1) = 2

For the right hand limit, the function occurs at x>-1

f+(x) = 2x-c

f+(-1) = 2(-1)-c

f+(-1) = -2-c

For the function f(x) to be continuous on the entire real line at x = -1, then

f-(-1) = f+(-1)

On equating both sides:

2 = -2-c

Add 2 to both sides

2+2 = -2-c+2

4 =-c

Multiply both sides by minus.

-(-c) = -4

c = -4

Hence the value of c so that f(x) is continuous on the entire real line is -4

6 0

3 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

4 years ago

5 square root 13^3. <br> A 13^2. <br> B 13^15 <br> C 13^5/3. <br> D 13^3/5

arlik [135]

Answer is D.

When you have an exponent that is a fraction, the denominator is the number you root the base by. In this case the exponent is 3/5, so you fifth root 13 cubed.

5 0

3 years ago

Quadrilateral ABCD has coordinates A (3,1) B (4,4) C (7,5) D (6,2). Quadrilateral ABCD is a ?

ASHA 777 [7]

The Quadrilateral ABCD with coordinates (3, 1), (4, 4), (7, 5), (6, 2) is a rhombus because its length and width are both square root of 10 units and adjacent sides are not perpendicular. This can be seen by plotting the points you can either plot it or by using programs.

I hope this helps, God bless, and have a great day!
Brainliest is always appreciated :)

3 0

3 years ago

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