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AysviL [449]
3 years ago
7

Consider two consecutive positive integers such that the square of the second integer added to 3 times the first is equal to 37.

Mathematics
1 answer:
wlad13 [49]3 years ago
4 0

Am I supposed to consider or solve?   Let's solve.

Let's call our integers x and x+1

square of the second integer added to 3 times the first is 37

(x+1)² + 3x = 37

x² + 2x + 1 + 3x - 37 = 0

x² + 5x - 36 = 0

(x+9)(x-4) = 0

Answer: x = -9 or x=4

Check:

x=-9

(-9+1)² + 3(-9) = 64 - 27 = 37 good

x=4

(4+1)² + 3(4) = 25 + 12 = 37 good

Two solutions.

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Answer:6300

Step-by-step explanation:

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3 years ago
The answer to this question
Sophie [7]

The answer to this question is D) 13.9 in.

Try using the Pythagorean theorem when solving to get the hypotenuse or other angles for a right triangle.

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3 years ago
Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:
Goshia [24]

Answer:

case 1 = 2592

case 2 =  729

case 1 + case 2 =  2916

(this is not a direct adition, because case 1 and case 2 have some shared elements)

Step-by-step explanation:

Case 1)

6 digits numbers that can be divided by 25.

For the first four positions, we can use any of the 6 given numbers.

For the last two positions, we have that the only numbers that can be divided by 25 are numbers that end in 25, 50, 75 or 100.

The only two that we can create with the numbers given are 25 and 75.

So for the fifth position we have 2 options, 2 or 7,

and for the last position we have only one option, 5.

Then the total number of combinations is:

C = 6*6*6*6*2*1 = 2592

case 2)

The even numbers are 2,4 and 6

the odd numbers are 3, 5 and 7.

For the even positions we can only use odd numbers, we have 3 even positions and 3 odd numbers, so the combinations are:

3*3*3

For the odd positions we can only use even numbers, we have 3 even numbers, so the number of combinations is:

3*3*3

we can put those two togheter and get that the total number of combinations is:

C = 3*3*3*3*3*3 = 3^6 = 729

If we want to calculate the combinations togheter, we need to discard the cases where we use 2 in the fourth position and 5 in the sixt position (because those numbers are already counted in case 1) so we have 2 numbers for the fifth position and 2 numbers for the sixt position

Then the number of combinations is

C = 3*3*3*3*2*2 = 324

Case 1 + case 2 = 324 + 2592 = 2916

4 0
3 years ago
1. Marisol is preparing a care package to send to her brother. The package will include a board game that weighs 4 pounds and se
Sladkaya [172]

The content of care package would be a board game and several snack packs.

Given is the weight of board game = 4 pounds.

The weight of one snack pack = \frac{1}{2} \;pounds = 0.5 \;pounds

The total weight would be less than 25 pounds.


Solving for part A :

Let 'x' represents the number of snack packs.

So the weight of all snack packs would be 0.5x pounds.

Now weight of care pack = weight of board game + weight of all snack packs.

weight of care pack = 4 + 0.5x

But total weight must be less than 25 pounds.

So the inequality would be: 4 + 0.5x < 25


Solving for part B :

Solving the inequality 4 + 0.5x < 25

⇒ 4 + 0.5x - 4 < 25 - 4

⇒ 0.5x < 21

⇒ x < 42


Solving for part C :

Since x < 42 and x is an integer number, so x = 41.

She can include maximum of 41 snack packs in the care package.

7 0
3 years ago
According to the Substitution Property of Equality: If y = -5 and 7x + y = 11, then _______ . Question 9 options: 7(-5) + y = 11
ycow [4]

Answer:

7x - 5 = 11

Step-by-step explanation:

To solve for x, substitute y = -5 into the equation 7x + y = 11.

This becomes 7x + (-5) = 11 or 7x - 5 = 11.

7 0
3 years ago
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