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drek231 [11]
2 years ago
10

What is the total surface area of the square pyramid whose net is shown?

Mathematics
1 answer:
ra1l [238]2 years ago
7 0

Answer:

1,600 sq. inches

Step-by-step explanation:

Area of a triangle:

bh/2

= (20 x 30)/2

= 600/2

= 300

Area of square:

s^2

= 20^2

= 400

Four triangles:

300 x 4

= 1200

Total:

1200 + 400

= 1600

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Can someone help please ASAP
Paul [167]

Answer:

4

Step-by-step explanation:

2 · 4 = 8

8 + 4 = 12

the value of x is 4

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2 years ago
Write the trinomial as a square of a binomial or as an expression opposite to a square of a binomial -42a+9a^2+49
Degger [83]

9(a^2+14/3)^2-14/3+49

9(a^2+14/3)^2+7

6 0
2 years ago
Police Chase: A speeder traveling 40 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off a
zubka84 [21]

Answer:

a. 18.34 s b. 327.92 m

Step-by-step explanation:

a. How long before the police car catches the speeder who continued traveling at 40 miles/hour

The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.

So, a =  (v - u)/t =  (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².

The distance moved by the police car in 10 s is gotten from

s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.

s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²

s = 0 m + 1/2 × 2.458 m/s² × 100 s²

s = 122.9 m

The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h

The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t

The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' =  17.88t' m where t' = time taken for police to catch up with speeder.

Since both distances are the same,

S' = S

17.88t' = 122.9 + 24.58t

Also, the time  taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t

t' = 10 + t

So, substituting t' into the equation, we have

17.88t' = 122.9 + 24.58t

17.88(10 + t) = 122.9 + 24.58t

178.8 + 17.88t = 122.9 + 24.58t

17.88t - 24.58t = 122.9 - 178.8

-6.7t = -55.9

t = -55.9/-6.7

t = 8.34 s

So, t' = 10 + t

t' = 10 + 8.34

t' = 18.34 s

So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour

b. how far before the police car catches the speeder who continued traveling at 40 miles/hour

Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m

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Latifah drew this obtuse, scalene triangle.
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Quadrilateral that is a parallelogram
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3 years ago
Find the degree of the monomial 7x^7 y^3
KengaRu [80]
I believe the answer is 7 but I’m 99% sure ,
Sorry if it’s incorrect
5 0
3 years ago
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