A group of adults and students went on a class trip to Washington, DC. The number of male students was 1 more than 7 times the number of adults. The number of female students was half the number of male students. If the total number of people who went on the trip is 82, find the numbers of male students and female students.

Answers

Answer 1

Answer: Given:
adults = x
male = 7x + 1
female = (7x + 1)/2
total number of people = 82

x + (7x + 1) + [(7x +1)/2] = 82
2x/2 + (14x + 2)/2 + (7x + 1)/2 = 82
2x + 14x + 2 + 7x + 1 = 82 * 2
23x + 3 = 164
23x = 164 - 3
23x = 161
x = 161/23
x = 7

adults = x = 7
males = 7x + 1 = 7(7) + 1 = 49 + 1 = 50
females = (7x+1)/2 = 50/2 = 25

7 + 50 + 25 = 82

Answer 2

Answer:

Answer:

50 males, 25 females

Step-by-step explanation:

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What adds to -2 and multiplies to 15

Answers

If you meant -15

the -5 and 3

3-5=-2
-3x5=-15

(-5) + 3 = -2(-5) × 3 = -15I can't think of one that multiplies to give positive 15 and also adds to give -2.

Can anyone plz help me plz this is the only question am asking ??:(

Answers

I think it is 1/7(3) but if I am wrong then sorry:)

What is the solution of the equation? 4(y 2) = 32 a. 6
b. –10
c. 10
d. 4?

Answers

32/4 = 8 <=> (y...2) = 8
if y/2 => x = 16
if yx2 => x = 4
if y + 2 => x = 6
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for his long distance phone service, Pablo pays an eight dollars monthly fee +6 cents per minute. Last month, Pablo's long-distance bill was $19.46. For how many minutes was Pablo billed?

Answers

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What is the product of (3+radical 5) and (3-radical 5)

Answers

(3 + $√(5)$ ) × (3 - $√(5)$ )
Take the special case
(a - b)×(a + b)=a² - b²

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3² - ( $√(5)$ )² {(√)² = the number itself}
9 - 5
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(3 + √((5)) + (3 - √(5))
3(3 - √(5)) + √(5)(3 - √(5)))
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1.solve the following equation for y 3x- 2y= 62. solve the following equation 1/3(6x-9)= 3/4(4x-8)
3. solve 3(2x-1)/4= x + 7/2
the 3(2x-1) is over the 4 and the x + 7 is over the 2 i really dont understand them

Answers

So,

For number one, there will be an infinite number of solutions, as there are two placeholders.
3x - 2y = 6
It seems we can use the intercept method.
3x - 2(0) = 6
3x = 6

Divide both sides by 3
x = 2

This is one solution: (2,0)

3(0) - 2y = 6
-2y = 6

Divide both sidees by -2
y = -3

This is another solution: (0,-3)
You should now be able to draw the line.
If you want intercept form for this equation, just manipulate the equation in order to get the y = mx + b form.

Subtract 3x from both sides
-2y = -3x + 6

Divide both sides by -2
$y = (3)/(2) x + 6$

For the second problem, simply manipulate the equation.
$(1)/(3) (6x - 9) = (3)/(4) (4x - 8)$

Distribute
$2x - 3 = 3x - 6$

Subtract 2x from both sides
$-3 = x - 6$

Add 6 to both sides
$3 = x$

For the third problem, do the same thing.
$(3(2x-1))/(4) = x + (7)/(2)$

We must first multiply both sides by 4.
$3(2x-1) = 4x + 14$

Distribute
$6x - 3 = 4x + 14$

Subtract 4x from both sides
$2x - 3 = 14$

Add 3 to both sides
$2x = 17$

Divide both sides by 2
$x = (17)/(2) \ or\ 8 (1)/(2)$

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