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3 years ago

I need the work shown not just the answers please

Mathematics

1 answer:

Yuliya22 [10]3 years ago

3 0

1.) To find the surface area, we want to find the area of each side, then add them all up together:
Area of base=lwA=(5)(5)A=25 cm²
Area of triangle=1/2bhA=(1/2)(5)(12.5)A=31.25 cm²
Since the 4 triangles have the same area, we can multiply the area we found by 4 and then add the area of the square:
SA=(4)(31.25) + 25SA=125+25SA=150 cm²

2.) <span>To find the volume use the formula:

V=(1/3)lwh
V=(1/3)(5)(5)(12.5)
V=104.167 cm^3</span>

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A hamster running in a wheel of radius 13 cm spins the wheel at 35 revolutions per minute.

crimeas [40]

One revolution is completed when a fixed point on the wheel travels a distance equal to the circumference of the wheel, which is 2π (13 cm) = 26π cm.

So we have

1 rev = 26π cm

1 rev = 2π rad

1 min = 60 s

(a) The angular velocity of the wheel is

(35 rev/min) * (2π rad/rev) * (1/60 min/s) = 7π/6 rad/s

or approximately 3.665 rad/s.

(b) The linear velocity is

(35 rev/min) * (26π cm/rev) * (1/60 min/s) = 91π/6 cm/s

or roughly 47.648 cm/s.

5 0

3 years ago

One side of a triangle is 2 times the second side. The third side is 15 feet longer than the second side. The perimeter of a

ch4aika [34]

Answer:

16, 8, 23

Step-by-step explanation:

2x+x+x+15=47

4x+15=47

4x=32

x=8

2x=16

x+15=23

7 0

2 years ago

HELLOOOOOooOOOOOooOOoooOOoo

Viefleur [7K]

Answer:

6. Multiply 2 by 6

7. 71 + 9(8) = 71 + 72 = 143

8. 52 + 3(4) = 12 + 52 = 64

9. 19.25 + 2.75(7) = 19.25 + 19.25 = 38.50

10. 2(0.6)^2 + 4(0.6)(5) = 0.72 + 12 = 12.72

Srry if they r wrong

6 0

2 years ago

Plzzzzzz help right answer gets brainly

GuDViN [60]

Answer:

3 * 3 * 3 * 3

=> $3^{4}$

If my answer helped, then kindly mark me as the brainliest!!

Thank You!!

6 0

3 years ago

Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]

alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

$\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]$

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)² = 3√6

Therefore, the unit vector is

$\frac{[7,2,-1]}{3\sqrt{6} }$

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

In conclusion, the two unit vectors are;

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

and

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

<em>Hope this helps!</em>

7 0

2 years ago