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2 years ago

Find two consecutive odd integers such that 62 more than the lesser is five times the greater.

Mathematics

2 answers:

lidiya [134]2 years ago

8 0

Idk really know srry

irakobra [83]2 years ago

7 0

So, let’s define 2 consecutive odd integers:
3 and 5
5 and 7
7 and 9
9 and 11

So, our integers would be written as “x” and “x+2”.

Our solution:

x+62 = 5(x+2)
x+62 = 5x + 10
x - 5x = 10 - 62
-4x = -52
x = -52/-4
x = 13

So our integers are 13 and 15.

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Alex has $20 and Ellen has $12 Alex saving three dollars per day and Ellen is saving five dollars per day after how many days wi

FinnZ [79.3K]

3x+20=5x+12
-3x -3x
20= 2x +12
-12 -12
8=2x
Divide both sides by two
4=x
It will take four days (this is how to solve algebraically)

3 0

3 years ago

When finding the 9th term in a geometric sequence with a common ratio of 2 and a first

Zarrin [17]

Answer:

Step-by-step explanation:

Nth term of a geometric sequence is given as:

$t_n = a r^{(n-1)}$ ...(1)

Plugging n = 9, a = 3 and r = 2 in the above equation, we find:

$t_9 = (3) (2)^{9-1}$

$t_9 = (3) (2)^{8}$ ... (2)

Comparing equations (1) & (2), we obtan:

(n - 1) = 8

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3 years ago

____________is the practice of putting students into specific curriculum groups based on their test scores and other factors.

Tresset [83]

Answer: Tracking is the practice of putting students into specific curriculum groups based on their test scores and other factors.

3 0

3 years ago

An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of

monitta

Answer:

As a fraction, the answer is exactly 86/969

In decimal form, the answer is approximately 0.08875

=====================================================

Work Shown:

The assumption is that no replacements are made for each selection.

5 red, 6 blue, 8 green

5+6+8 = 19 total

A = P(3 red) = (5/19)*(4/18)*(3/17) = 10/969

B = P(3 blue) = (6/19)*(5/18)*(4/17) = 20/969

C = P(3 green) = (8/19)*(7/18)*(6/17) = 56/969

D = P(3 all same color)

D = A+B+C

D = 10/969 + 20/969 + 56/969

D = (10+20+56)/969

D = 86/969

D = 0.08875

6 0

3 years ago

Suppose an airline policy states that all baggage must be box shaped with a sum of length, width, and height not exceeding 174 i

Leya [2.2K]

Answer:

The square-based box with the greatest volume under the condition that the sum of length, width, and heigth does not exceed 174 in is a cube with each edge of 58 in and a volume of $195112 in^{3}$

Step-by-step explanation:

For this problem we have two constraints, that are as follows:

1) Sum of length, width, and heigth not exceeding 174 in

2) Lenght and width have the same measure (square-based box)

We know that volume is equal to the product of all three edges, and with the two conditions into account we have the next function:

$V=(w^{2})(174-2w)\\V=174w^{2}-2w^{3}$

The interval of interest of the objective function is [0, 87]

This problem requieres that we maximize the function that defines the volume. We start calculating the derivative of the function, wich is:

$V'=348w-6w^{2} \\V'=(348-6w)(w)$

We need to remember that the derivative of a function represents the slope of said function at a given point. The maximum value of the function will have a slope equal to zero.

So we find the value in wich the derivative equals zero:

$0=(348-6w)(w)\\w_1=0\\w_2=348/6=58$

The first value ( $w=0$ ) will leave us with a 'height-only box', so the answer must be $w=58 in$

The value is between the interval of interest.

And, once we solve for the constraints, we have that:

Lenght = Width = Heigth = 58 in

Volume = $195112 in^{2}$

8 0

3 years ago