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mel-nik [20]
3 years ago
6

Find the surface area of the right triangular prism (above) using its net (below).

Mathematics
1 answer:
lakkis [162]3 years ago
7 0

Answer:

144 units²

Step-by-step explanation:

The net of the right triangular prism consists of 3 rectangles and 2 equal triangles

Let's solve for the area of each:

✔️Area of rectangle 1 = L*W

L = 11

W = 3

Area of rectangle 1 = 11*3 = 33 units²

✔️Area of rectangle 2 = L*W

L = 11

W = 4

Area of rectangle 2 = 11*4 = 44 units²

✔️Area of rectangle 3 = L*W

L = 11

W = 5

Area of rectangle 3 = 11*5 = 55 units²

✔️Area of the two triangles = 2(½*base*height)

base = 4

height = 3

Area of the two traingles = 2(½*4*3)

= 12 units²

✔️Surface area of the right triangle = area of rectangle 1 + area of rectangle 2 + area of rectangle 3 + area of the two triangles

= 33 + 44 + 55 + 12

= 144 units²

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A linear function is defined by the equation f(x)=-2(x+5). The table below represents another linear function ,g(x). What is the
astra-53 [7]

Answer:

b. 3

Step-by-step explanation:

find equation of g(x):

(-16-(-10))/(-3-(-1))=-6/-2=3

apply a point

-16=3(-3)+n

-16=-9+n

n=-7

g(x)=3x-7

find equation of f(x):

f(x)=-2(x+5)

simplify

f(x)=-2x-10

subtract y intercepts

-7-10=3

3

6 0
3 years ago
A spinner can land on either red, blue, green, or yellow (the wedges are the same size). you spin twice. what is the probability
melisa1 [442]
There are four (4) colors to choose from. The number of sample space is equal to 4. With the condition given in this item, the probability of landing on blue on the first spin is equal to 1/4. Similarly, the probability of landing on yellow on the second spin is 1/4. 

            P = (1/4)(1/4)
  
              P = 1/16

Answer: 1/16
5 0
4 years ago
Brad jumps 4 feet forward and then 2 feet backward.How many full sets must he jump to each 16 feet
julsineya [31]
8 sets, she’s only traveling 2 feet forward per each set 2*x=16 x=8
6 0
2 years ago
What is the x-intercept of the graph 0f 27= -3x2?
jok3333 [9.3K]

Answer:

x=±√−9

No real solutions.

Step-by-step explanation:

Step 1: Add 3x^2 to both sides.

27+3x2=−3x2+3x2

3x2+27=0

Step 2: Subtract 27 from both sides.

3x2+27−27=0−27

3x2=−27

Step 3: Divide both sides by 3.

3x2 /3 = −27 /3

x2=−9

Step 4: Take square root.

x=±√−9

5 0
3 years ago
Two airplanes leave an airport at the same time, the first headed due north and the second at a bearing of N42^ E . At 2:00PM, t
Leno4ka [110]

Answer:

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.

c² = a² + b² - 2abcosC            --------------(i)

where;

b  = 312miles

a = 487miles

C = 42°

Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

c² = (237169) + (97344) - 303888cos(42)

c² = (237169) + (97344) - 303888(0.7431)

c² = 334513 - 225819.1728

c² = 108693.8272

<em>Take the square root of both sides</em>

√c² = √108693.8272

c = 329.69

c ≅ 330miles

Therefore, the two airplanes are far apart by 330miles

3 0
3 years ago
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