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vaieri [72.5K]
2 years ago
9

Find the SURFACE AREA of the shape below.

Mathematics
1 answer:
dedylja [7]2 years ago
3 0

Answer:

292 CM

Step-by-step explanation:

8x7 = 56 (2 such sides)

6x8 = 48 (2 such sides)

7x6 = 42 (2 such sides)

56(2) + 48(2) + 42(2)=

292

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Suppose on the first day, 80 adult, 120 student, and 20 senior tickets were sold with a total of $1,460 in ticket sales. On the
Debora [2.8K]
<h2>✏<u>Answer:</u></h2>

She spend $1,238 a month

45 people can get senior tickets

110 students

<h2>#CarryOnLearning</h2>

4 0
2 years ago
If 12% of a number is 24 what is the number Pls Answer fast
Nadya [2.5K]

Answer:

200

Step-by-step explanation:

12% is equal to 12/100

I would set up something like

12/100= 24/x

to get from 12 to 24, you have to multiply by 2.

do the same for 100 so 100 x 2= 200

5 0
3 years ago
Read 2 more answers
PLEASE HELP PLEASEEE!!!!
shtirl [24]
In first diameter=8,In second diameter=2/3,
6 0
3 years ago
Given: <br> PQ<br> ⊥<br> QR<br> , PR=20,<br> SR=11, QS=5<br> Find: The value of PS.
dlinn [17]

Answer:

The value of the side PS is 26 approx.

Step-by-step explanation:

In this question we have two right triangles. Triangle PQR and Triangle PQS.

Where S is some point on the line segment QR.

Given:

PR = 20

SR = 11

QS = 5

We know that QR = QS + SR

QR = 11 + 5

QR = 16

Now triangle PQR has one unknown side PQ which in its base.

Finding PQ:

Using Pythagoras theorem for the right angled triangle PQR.

PR² = PQ² + QR²

PQ = √(PR² - QR²)

PQ = √(20²+16²)

PQ = √656

PQ = 4√41

Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.

Finding PS:

Using Pythagoras theorem, we have:

PS² = PQ² + QS²

PS² = 656 + 25

PS² = 681

PS = 26.09

PS = 26

7 0
3 years ago
The graph below shows the function f(x)=x-3/x^2-2x-3 which statement is true
Ugo [173]

Answer:

The correct option is A.

Step-by-step explanation:

Domain:

The expression in the denominator is x^2-2x-3

x² - 2x-3 ≠0

-3 = +1 -4

(x²-2x+1)-4 ≠0

(x²-2x+1)=(x-1)²

(x-1)² - (2)² ≠0

∴a²-b² =(a-b)(a+b)

(x-1-2)(x-1+2) ≠0

(x-3)(x+1) ≠0

x≠3 for all x≠ -1

So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong

Asymptote:

x-3/x^2-2x-3

We know that denominator is equal to (x-3)(x+1)

x-3/(x-3)(x+1)

x-3 will be cancelled out by x-3

1/x+1

We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....

4 0
3 years ago
Read 2 more answers
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