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2 years ago

3. If 3, x, y, 18, are in arithmetic progression, (A, P) find the value of x and y

Mathematics

1 answer:

Luba_88 [7]2 years ago

5 0

Answer:

Below in bold

Step-by-step explanation:

The sequence is:

3, x, y, 18

If this is an A P then

x - 3 = y - x

2x - y = 3 (A) and

y - x = 18 - y

2y - x = 18 (B)

Multiply (A) by 2:

4x - 2y = 6 (C)

Adding B and C:

3x = 24

x = 8.

and

2y - 8 = 18

2y = 26

y = 13.

So x = 8 and y = 13.

b) ar + ar^2 = 6ar^3 where a = first term and r = common ratio

Divide by a:

r + r^2 = 6r^3

6r^3 - r^2 - r = 0

r(6r^2 - r - 1) = 0

r(3r + 1)(2r - 1) = 0

So the 2 possible values of r

= -1/3 and 1/2.

i) The common ratio is positive so it must be 1/2.

Second term ar = 8

1/2 a = 8

a = 16.

So the first 6 terms are:

16, 8, 4, 2, 1, 1/2.

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<h3>What are sets of numbers?</h3>

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1 year ago

A line with a slope of -3 that passes through the point (-2, 6)

scZoUnD [109]

I believe the equation is y = -3x + 0

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2 years ago

Factorise 2x^2 - 3x -2 < 0

Galina-37 [17]

Step 1: Factor $2 x^{2} - 3x -2$
1. Multiply 2 by -2, which is -4.
2. Ask: Which two numbers add up to -3 and multiply to -4?
3. Answer: 1 and -4
4. Rewrite $-3x$ as the sum of $x$ and $-4x$

$2 x^{2} +x-4x-2\ \textless \ 0$

Step 2: Factor out common terms in the first two terms, then in the last two terms.

 $x(2x+1)-2(2x+1)\ \textless \ 0$

Step 3: Factor out the common term $2x+1$

$(2x+1)(x-2)\ \textless \ 0$

Step 4: Solve for $x$

1. Ask: When will $(2x+1)(x-2)$ equal zero?
2. Answer: When $2x + 1 = 0$ or $x-2=0$
3. Solve each of the 2 equations above:

 $x=- \frac{1}{2} ,2$

Step 5: From the values of $x$ above, we have these 3 intervals to test.
x = < -1/2
-1/2 < x < 2
x > 2

Step 6: Pick a test point for each interval

For the interval $x\ \textless \ - \frac{1}{2} 
$
Lets pick $x=-1.$ Then, $2(-1) ^{2} -3 * -1 -2 \ \textless \ 0$
After simplifying, we get $3\ \textless \ 0$ , Which is false.
Drop this interval.

For this interval $- \frac{1}{2} \ \textless \ x\ \textless \ 2$
Lets pick $x=0$ . Then, $2* 0^{2} - 3 * 0-2\ \textless \ 0$ . After simplifying, we get $-2\ \textless \ 0,$ which is true. Keep this interval.

For the interval $x\ \textgreater \ 2$

Lets pick $x = 3.$ Then, $2 * 3 ^{2} -3*3-2\ \textless \ 0.$ After simplifying, we get $7\ \textless \ 0$ , Which is false. Drop this interval.

.Step 7: Therefore,
$- \frac{1}{2} \ \textless \ x\ \textless \ 2$

Done! :)

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