PLSSSSS HELPPPPPPP ASAPPPPPP!!!! I WILLL GIVE BRAINLIEST!!!!

WILL GIVE BRAINLIEST<br> FIND VOLUME<br> 60 is the area of the base

scoundrel [369]

Hey so yeah this can be a challenging problem, Vol (V) is much easier to solve than surface area (SA), but I'll show you how it's done, my friend.
First of all, V (prism) = area of base (B) × h
We know that the height (h) of this prism is given, which we'll need later on: h = 5 ft, and area of base (B) is given as 60 ft2
So V = 60 ft2 × 5 ft = 300 ft3

So now for the hard part... how to calculate the SA of this prism

IF YOU DON'T NEED SA FOR THIS TYPE OF PROBLEM, DO NOT PROCEED!!!
[VERY DETAILED]

means solving the dimensions (sides) of that pesky polygon base (B) or lid.
The most important things about polygons are:
1) is it regular (same angle ° and side length)??
2) How many sides or angles??

This has to be regular, because they give you no other info so it has to be, in order to solve. And then it has 5 sides and angles = regular pentagon. ("penta" means 5).

Now there are 360° in any circle, so:
take the central angles of where the sides meet at the center forming triangles (see drawing above), each of those (5) central <'s = 360/5 = 72°
Now the apex of each of these triangles = 72°
but with our 5 triangles, we need to find the height of each triangle -- which is the midpoint of the side (base (b) of triangle), and h is perpendicular to this b. By bisecting that apex angle of 72, it forms 2 equal right triangles of
72/2 = 36°. So each right triangle has 36, 90, and?? 180-36-90 = 90-36 = 54°
let's call the base (b) = 1 side of pentagon
= side (s)
Therefore (see 2nd image drawn above) tangent (tan) of € = opposite/adjacent, or
tan (54) = h÷1/2b --> h = [tan (54)]×(1/2)b
1/2b = h/[tan (54)] = h/1.38

Also the area of each of the larger 5 triangles A(t) = 1/2b×h, and that area A(t) × 5 = area of whole pentagon base A(B)
So now after all that... our A(B) given at beginning = 60ft2. let's put it all together:
1/2b = h/[tan (54)] = h/1.38
A(t) = 1/2b×h, and A(B) = 5×A(t)
which means that A(B) = 5×(1/2b×h)
AND since the other calculation shows that 1/2b = h/1.38, plug that value into the A(B) formula...
.
.
.

now that we have height (h) of each triangle, we can go back to our tan equation for the triangle: tan (54) = 1.38 = h/(1/2b)
--> 1/2×b = h/1.38 --> base (b) = 2h/1.38
b = 2(4.06 ft)/1.38 = 8.12 ft/1.38 = 5.90 ft, for which b us also the width of each rectangular side panel (w)
length (l) of these sides was given as height of the whole prism = 5 ft

NOW FINALLY... THE FINAL SURFACE AREA OF THE PRISM (SA) = (2×A(B)) + (5×rectangular side)
SA = (2×60 ft2) + (5×(l×w)) = 120 ft2 + 5×5ft×5.9ft
SA = 120 ft2 + 147.62 ft2 = 267.62 ft2

4 0

3 years ago

Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road-oriented at an angle of 25

kap26 [50]

Answer:

6.31 mi

Step-by-step explanation:

The diagram below explains the solution better.

From the diagram,

C = starting point of the race.

A = end of the first part of the race.

B = end of the race.

Using Cosine rule, we can find the straight-line distance between the starting point and the end of the race.

Cosine rule states that:

$a^2 = b^2 + c^2 - 2bc[cos(A)]$

where A = angle A = <A

Given that

b = 5.2 miles

c = 2.0 miles

<A = 115° (from the diagram)

Hence,

$a^2 = 5.2^2 + 2.0^2 - 2*5.2*2.0[cos(115)]\\\\a^2 = 27.04 + 4 - 20.8[cos(115)]\\\\a^2 = 31.04 + 8.79\\\\a^2 = 39.83\\\\a = \sqrt{39.83}\\ \\a = 6.31 mi$