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3 years ago

How much must be added to each of the three numbers 1, 11, and 23 so that together they form a geometric progression?

Mathematics

1 answer:

Sliva [168]3 years ago

5 0

Answer:

Step-by-step explanation:

Let x be unknown number which should be added to numbers 1, 11, 23 to get geometric progression. Then numbers 1 + x, 11 + x, 23 + x are first three terms of geometric progression.

Hence,

$b_1=1+x\\ \\b_2=11+x\\ \\b_3=23+x$

and

$b_2=b_1\cdot q\Rightarrow 11+x=(1+x)q\\ \\b_3=b_2\cdot q\Rightarrow 23+x=(11+x)q$

Express q:

$q=\dfrac{11+x}{1+x}=\dfrac{23+x}{11+x}$

Solve this equation. Cross multiply:

$(11+x)^2=(1+x)(23+x)\\ \\121+22x+x^2=23+x+23x+x^2\\ \\121+22x=23+24x\\ \\22x-24x=23-121\\ \\-2x=-98\\ \\2x=98\\ \\x=49$

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Solnce55 [7]

12,989 is the answer i believe

3 0

3 years ago

Help !

aivan3 [116]

Answer:

a. p(orange) = 5/14

b. p(green) = 3/14

c. p(red) = 1/7

d. p(brown) = 2/7

e. p(brown or red) = 3/7

Step-by-step explanation:

1. You have a 14 pencils. Two pencils are red, 5 pencils are orange, 3 pencils are green and 4 pencils are brown.

p(color) = (number of pencils of that color)/(total number of pencils)

p(color) = (number of pencils of that color)/14

a. If a pencil is picked at random, what is the probability that the pencil

will be orange?

p(orange) = 5/14

b. If a pencil is picked at random, what is the probability that the pencil

will be green?

p(green) = 3/14

c. If a pencil is picked at random, what is the probability that the pencil will be red?

p(red) = 2/14 = 1/7

d. If a pencil is picked at random, what is the probability that the pencil

will be brown?

p(brown) = 4/14 = 2/7

e. If a pencil is picked at random, what is the probability that the pencil

will be brown or red?

brown: 4

red: 2

brown or red: 4 + 2

p(brown or red) = 6/14 = 3/7

3 0

3 years ago

Brianna has x nickels and y pennies. She has no less than 20 coins worth no more

Bond [772]

Answer:

$y \ge 20 - x$

$y \le 80 -5x$

$(x,y) = (15,5)$

Step-by-step explanation:

Given

$Nickels = x$

$Pennies = y$

$Amount = \$0.80$ maximum

$Coins = 20$ minimum

Required

Solve graphically

First, we need to determine the inequalities of the system.

For number of coins, we have:

$x\ +\ y\ge 20$ because the number of coins is not less than 20

For the worth of coins, we have:

$0.05x\ +\ 0.01y\ \le0.80$ because the worth of coins is not more than 0.80

So, we have the following equations:

$x\ +\ y\ge 20$

$0.05x\ +\ 0.01y\ \le0.80$

Make y the subject in both cases:

$y \ge 20 - x$

$0.01y \le 0.80 - 0.05x$

Divide through by 0.01

$\frac{0.01y}{0.01} \le \frac{0.80}{0.01} -\frac{ 0.05x}{0.01}$

$y \le \frac{0.80}{0.01} -\frac{ 0.05x}{0.01}$

$y \le 80 -5x$

The resulting inequalities are:

$y \ge 20 - x$

$y \le 80 -5x$

The two inequalities are plotted on the graph as shown in the attachment.

$y \ge 20 - x$ --- Blue

$y \ge 80 -5x$ --- Green

Point A on the attachment are possible solutions

At A:

$(x,y) = (15,5)$

4 0

2 years ago

Solve the following inequality: 3-5x-x^2>=0

Dennis_Churaev [7]

Answer:

-5/2+-1/2√37≤x≤-5/2+1/2√37

Step-by-step explanation:

Step 1: Find the critical points

-x^2-5x+3=0

For this equation: a=-1, b=-5, c=3

−1x^2+−5x+3=0

x=−b±√b2−4ac/2a

x=−(−5)±√(−5)2−4(−1)(3)/2(-1)

x=5±√37 /−2

x=-5/2+1/2√37

Step 2: Check intervals in between critical points

x≤-5/2+1/2 √37 (Doesn't work in original inequality)

-5/2+-1/2√37≤x≤-5/2+1/2√37 (Works in original inequality)

x≥-5/2+1/2 √37 (Doesn't work in original inequality)

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3 years ago

Find the measure of angle ABC

victus00 [196]

First we need to find x so 6x+8x+6x= 180
20x=180 and x= 9

So A= 6*9= 54
B= 8*9= 72
C= 6*9= 54

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2 years ago