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3 years ago

3a/(a+1)^2 Give the equivalent numerator if the denominator is (a + 1)³.

Mathematics

2 answers:

liberstina [14]3 years ago

6 0

Answer:

3a(a+1)

Step-by-step explanation: for those who were confused on the other asnwer he gave a different veriation of this answer. however it is correct.

Brainliest plsss

zlopas [31]3 years ago

3 0

Given 3a/(a+1)^2
To make the denominator a cube, you would have to multiply by 1 or (a+1)/(a+1)
yielding 3a(a+1)/(a+1)^3
(3a^2 + 3a) is the equivalent numerator

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3x^2+6x-9=0 complete the square

PolarNik [594]

You could simplify this work by factoring "3" out of all four terms, as follows:

3(x^2 + 2x - 3) =3(0) = 0

Hold the 3 for later re-insertion. Focus on "completing the square" of x^2 + 2x - 3.

1. Take the coefficient (2) of x and halve it: 2 divided by 2 is 1
2. Square this result: 1^2 = 1
3. Add this result (1) to x^2 + 2x, holding the "-3" for later:
x^2 +2x
4 Subtract (1) from x^2 + 2x + 1: x^2 + 2x + 1 -3 -1 = 0,
or x^2 + 2x + 1 - 4 = 0
5. Simplify, remembering that x^2 + 2x + 1 is a perfect square:

(x+1)^2 - 4 = 0

We have "completed the square." We can stop here. or, we could solve for x: one way would be to factor the left side:

[(x+1)-2][(x+1)+2]=0 The solutions would then be:

x+1-2=0=> x-1=0, or x=1, and
x+1 +2 = 0 => x+3=0, or x=-3. (you were not asked to do this).

3 0

3 years ago

A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y equals = 100 100 minus −x squa

masha68 [24]

Answer:

Step-by-step explanation:

Given that a rectangle is constructed with its base on the x axis and two of its vertices on the parabola

$y=100-x^2$

This parabola has vertex at (0,100) and symmetrical about y axis.

Any general point above x axis can be written as (a,b) (-a,b) since symmetrical about yaxis.

Hence coordinates of any rectangle are

$(a,0) (-a,0), (a, 100-a^2), (-a, 100-a^2)$

Length of rectangle = 2a and width = $100-a^2$

Area of rectangle = lw = $2a(100-a^2)=200a-400a^3$

To find max area, use derivative test.

$A' = 200-800a^2\\A"=-1600a$

Hence maxima when first derivative =0

i.e. when a =2

Thus we find dimensions of the rectangle are l =4 and w = 96

Maximum area = $4(96) = 384$

5 0

3 years ago

Sophie drew a scale drawing of a house and its lot. The back patio is 3 centimeters wide in the drawing. The actual patio is 18

wel

Answer: 45?

Step-by-step explanation: Sorry If I'm wrong, I haven't learned this yet but I tried my best.

4 0

3 years ago

First you to find the worksheet and download it<br> plase I need help

Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = $3^{x + 2}$

The asymptote is given as x → -∞, f(x) = $3^{x + 2}$ → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = $3^{0 + 2}$ = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = $5^{1 - x}$

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = $5^{1 - 0}$ = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = $2^{x + 3}$ - 1

The asymptote is $2^{x + 3}$ → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) = $2^{0 + 3}$ - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, $2^{x + 3}$ - 1 = 0

$2^{x + 3}$ = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) $f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2$

The asymptote is $\left (\dfrac{1}{2} \right)^{x - 2}$ → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = $f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6$

The y-intercept is (0, 6)

7 0

2 years ago

To celebrate there 125 anniversary, a company in Germany produced 125 very expensive teddy bears. The bears,known as the "125 ka

liq [111]

Answer:

The cost of one bear is <u>$47,000</u>.

Step-by-step explanation:

The question is incomplete and the data is missing so the complete question with data is below with attached files:

To celebrate there 125 anniversary, a company in Germany produced 125 very expensive teddy bears. The bears,known as the "125 karat teddy bears", are made mohair,silk and gold thread and have diamond and sapphire eyes.

The chart shows the approximate cost of " 125 teddy bears". Based on the pattern, how much does one bear cost?

Now, to get the cost of one bear.

So, we solve it by using unitary method:

If 4 bears cost = $188,000.

Then 1 bear cost = $188000\div 4=47000$

Thus, cost of one bear is $47,000.

Now, we check it by using another pattern of the data:

If 7 bears cost = $329,000.

Then 1 bear cost = $329000\div 7=47000.$

Hence, the cost of one bear is $47,000.

Therefore, the cost of one bear is $47,000.

8 0

3 years ago