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ladessa [460]
3 years ago
6

If you could help I would really appreciate it, but if not that’s fine. Thank you.

Mathematics
1 answer:
drek231 [11]3 years ago
3 0

Answer:

c po ang sagot jan parehas tayo

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PLSS HELP ME FIND THE SURFACE AREA OF THE TRIANGULAR PRISM <br><br> THXXSSSS AND HURRY PLS
aleksley [76]

Answer:

108m^2

Step-by-step explanation:

surface area of any shape: sum of the area of its sides.

area of cross section x 2 = 4 x 3 12 m^2

area of slope:  5m x 8m = 40m^2

area of base: 8 X 3 = 24 m^2

area of back: 8m x 4m = 32m^2

sum = surface area = 108m^2

Please mark brainiest

5 0
3 years ago
How do you solve f(x) = x2 + x - 20?​
Darina [25.2K]

\huge \bf༆ Answer ༄

Let's solve ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: {x}^{2}  + x - 20

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: {x}^{2}  + 5x - 4x - 20

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x(x + 5) - 4(x + 5)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(x + 5)(x -  4)

It's factorized now ~ And if you want to find the zeros then equate the expression with 0.

And you will get ;

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x =  - 5 \:  \: and \:  \: 4

4 0
2 years ago
What is the distance between point P1 and P2?
julia-pushkina [17]
The rise is 1 and the run is 7
5 0
3 years ago
Write an<br> explicit formula for an, the nth term of the sequence 24, 28, 32, ....
noname [10]

Answer:

Given the sequence 20, 24, 28, 32, 36, . . . find the 60th term.

------

nth term pattern: a(n) = 20 + (n-1)4

60th term: a(60) = 20 + 59*4 = 256

Step-by-step explanation:

3 0
3 years ago
Given cosα = −3/5, 180 &lt; α &lt; 270, and sinβ = 12/13, 90 &lt; β &lt; 180
torisob [31]

I got

-  \frac{6 3}{65}

What we know

cos a=-3/5.

sin b=12/13

Angle A interval are between 180 and 270 or third quadrant

Angle B quadrant is between 90 and 180 or second quadrant.

What we need to find

Cos(b)

Cos(a)

What we are going to apply

Sum and Difference Formulas

Basics Sine and Cosines Identies.

1. Let write out the cos(a-b) formula.

\cos(a - b)  =  \cos(a)  \cos(b)  +  \sin(a)  \sin(b)

2. Use the interval it gave us.

According to the given, Angle B must between in second quadrant.

Since sin is opposite/hypotenuse and we are given a sin b=12/13. We. are going to set up an equation using the pythagorean theorem.

.

{12}^{2}  +  {y}^{2}  =  {13}^{2}

144 +  {y}^{2}  = 169

25 =  {y}^{2}

y = 5

so our adjacent side is 5.

Cosine is adjacent/hypotenuse so our cos b=5/13.

Using the interval it gave us, Angle a must be in the third quadrant. Since cos is adjacent/hypotenuse and we are given cos a=-3/5. We are going to set up an equation using pythagorean theorem,

.

( - 3) {}^{2}  +  {x}^{2}  =  {5}^{2}

9 +  {x}^{2}  = 25

{x}^{2}  = 16

x = 4

so our opposite side is 4. sin =Opposite/Hypotenuse so our sin a =4/5.Sin is negative in the third quadrant so

sin a =-4/5.

Now use cosine difference formula

-  \frac{3}{5}  \times  \frac{5}{13}  +   - \frac {4}{5}  \times  \frac{12}{13}

- \frac{15}{65} + (  - \frac{48}{65}  )

-  \frac{63}{65}

Hope this helps

6 0
3 years ago
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