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avanturin [10]
3 years ago
14

Help ASAP PLEASE AND SHOW HOW YOU GOT THE ANSWER

Mathematics
1 answer:
Kruka [31]3 years ago
3 0
480 units, because
Y × X = 65
X × Z = 104
Y × Z = 40


104 \times 2 > 208
40 \times 2 = 80
65 \times 2 = 130
You multiple each answer by 2 because there are 2 sides(of the same size) on the shape.
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Find the 73rd term of the arithmetic sequence -28, -41, -54, ...
PolarNik [594]

Answer:

the 73rd term of the arithmetic sequence is -964.

Step-by-step explanation:

The common difference in this arithmetic sequence is 13.  This is obtained by subtracting 28 from 41.  The first term is a(1) = -28.

The arithmetic sequence formula is a(n) = a(1) + d(n - 1), where d is the common difference and -28 the first term.

Then, in this case, a(n) = -28 - 13(n - 1), and

a(73) = -28 - 13(73 - 1) = -964

3 0
2 years ago
Please answer (*´∇`*)
garri49 [273]
Answer:
D. 26

Explanation:
10% = 6.5
6.5 x 4 = 26
6 0
3 years ago
Solve for X.<br> 4(2x + 5) + 2x + 3 = -11
S_A_V [24]
We can begin by rewriting the problem:

4(2x + 5) + 2x + 3 = -11

Let’s solve in steps.

1. Distribute

8x + 20 + 2x + 3 = -11

2. Combine like terms

10x + 23 = -11

3. Subtract 23 from both sides

10x + (23 - 23) = -11 - 23

10x = -34

4. Divide both sides by 10

10 / 10 = 1 (or x)

-34 / 10 = -3.4

Therefore, x = -3.4.
3 0
1 year ago
Read 2 more answers
In the game of blackjack played with one​ deck, a player is initially dealt 2 different cards from the 52 different cards in the
Volgvan

Answer:

P =4.83\%

Step-by-step explanation:

First we calculate the number of possible ways to select 2 cards an ace and a card of 10 points.

There are 4 ace in the deck

There are 16 cards of 10 points in the deck

To make this calculation we use the formula of combinations

nCr=\frac{n!}{r!(n-r)!}

Where n is the total number of letters and r are chosen from them

The number of ways to choose 1 As is:

4C1 = 4

The number of ways to choose a 10-point letter is:

16C1 = 16

Therefore, the number of ways to choose an Ace and a 10-point card is:

4C1 * 16C1 = 4 * 16 = 64

Now the number of ways to choose any 2 cards from a deck of 52 cards is:

52C2 =\frac{52!}{2!(52-2)!}

52C2 = 1326

Therefore, the probability of obtaining an "blackjack" is:

P = \frac{4C1 * 16C1}{52C2}

P = \frac{64}{1326}

P = \frac{32}{663}

P =0.0483

P =4.83\%

4 0
3 years ago
Read 2 more answers
For 3 consecutive odd integers, the result of adding the smallest integer to four times the largest integer is 121. What are the
irakobra [83]

Answer: 21, 23, 25

Step-by-step explanation:

Let the integers be y, y+2 and y+4.

From the question, we are informed that the result of adding the smallest integer to four times the largest integer is 121. This can be mathematically expressed as:

y + 4(y+4) = 121

y + 4y + 16 = 121

5y + 16 = 121

5y = 121-16

5y = 105

y = 105/5

y = 21

Since y = 21,

y + 2 = 21 + 2 = 23

y + 4 = 21 + 4 = 25

The integers are 21,23 and 25

4 0
3 years ago
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