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3 years ago

Can someone please help me with number 30?

Mathematics

1 answer:

kenny6666 [7]3 years ago

5 0

This is an application of the pythagorean theorem.
c^2 = h^2 + b^2
c = 190
b = 102.5
h = ???

190^2 = h^2 + 102.5^2
16100 = h^2 + 10506.25
36100 - 10506.25 = h^2
25593.75 = h^2
sqrt(25593.75) = h
159.98 = h

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All 203 students at Polk school eat lunch at the same time One day 19 students were absent if 8 students sit at each table yhw l

lisov135 [29]

Answer:

23 tables

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the sum of two consecutive even integers divided by four is 189.5?

mixas84 [53]

Let us denote the smaller even integer by $N$ .
Then, the next even integer must be $N+2$ .

To verify, set $N$ as any even number, say $174$ .
Then, the next even integer must be 2 more than this number, i.e. $174+2=176$ .

The sum of these two consecutive even integers can be expressed as:
                                    $N+(N+2)=2N+2$

We are told that this sum divided by four is 189.5, i.e.
                                     $(2N+2)\div4=189.5$

Multiplying both sides by 4 will reverse the division:
     $(2N+2)\div4\times4=189.5\times4$
     $(2N+2)\times1=758$

Subtracting 2 from each side will reverse the addition:
     $2N+2-2=758-2$
     $2N=756$

Dividing both sides by 2 will reverse the multiplication:
     $2N\div2=756\div2$
     $N=378$

Therefore, the two consecutive integers must be 378 and 380.

Check this:
     $(378+380)\div4=758\div4=189.5$

3 0

3 years ago

three adults and four children must pay $107. two adults and three children must pay $76. find the price of the adult ticks and

jenyasd209 [6]

Assigning "a" as cost of adult tickets and "c" the cost of children's tickets.

3a + 4c = 107
2a + 3c = 76

elimination method
2(3a + 4c = 107)
-3(2a + 3c = 76)

6a + 8c = 214
-6a - 9c = -228
----------------------
0 - c = - 14
c = 14

Then 2a + 3(14) = 76
2a = 76 - 42
2a = 34
a = 34/2
a = 17

8 0

3 years ago

Solve for c: c+8=-11c

UNO [17]

Answer:

c= -2/3

Step-by-step explanation:

c+8 = -11c

C = -11c -8

c+11c = -8

12c = -8

divide both sides by 12

c= -2/3

7 0

3 years ago

Read 2 more answers

Q‒1. [5×4 marks] a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6? (150) b) How many three-

amid [387]

Answer:

a) 294

b) 180

c) 75

d) 174

e) 105

Step-by-step explanation:

I assume that for each problem, the first digit can't be 0.

a) There are 6 digits that can be first, 7 digits that can be second, and 7 digits that can be third.

6×7×7 = 294

b) This time, no digit can be used twice, so there are 6 digits that can be first, 6 digits that can be second, and 5 digits that can be third.

6×6×5 = 180

c) Again, each digit can only be used once, but this time, the last digit must be odd.

If only the last digit is odd, there are 3×3×3 = 27 possible numbers.

If the first and last digits are odd, there are 3×4×2 = 24 possible numbers.

If the second and last digits are odd, there are 3×3×2 = 18 possible numbers.

If all three digits are odd, there are 3×2×1 = 6 possible numbers.

The total is 27 + 24 + 18 + 6 = 75.

d) If the first digit is 3, and the second digit is 3, there are 1×1×6 = 6 possible numbers.

If the first digit is 3, and the second digit is greater than 3, there are 1×3×7 = 21 possible numbers.

If the first digit is greater than 3, there are 3×7×7 = 147 numbers.

The total is 6 + 21 + 147 = 174.

e) If the first digit is 3, and the second digit is greater than 3, then there are 1×3×5 = 15 possible numbers.

If the second digit is greater than 3, there are 3×6×5 = 90 possible numbers.

The total is 15 + 90 = 105.

6 0

3 years ago