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2 years ago

Expand and combine like terms. (4b^2+ 3) (4b^2- 3) = 1 SEE ANSWER

Mathematics

1 answer:

krok68 [10]2 years ago

8 0

Answer:

Step-by-step explanation:

(4b^2+3)(4b^2-3)

since it is the expanded form of formula a^2-b^2=(a+b)(a-b) we can write,

=(4b^2)^2-(3)^3

=16b^4-9

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P over 2 minus 5 when p = 14

bezimeni [28]

Answer: (14/2)-5=2

Step-by-step explanation:

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3 years ago

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The radius of a circle is 10 cm. Find its circumference in terms of pi

sergejj [24]

Answer:

20 pi is the correct answer

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2 years ago

PLEASE HELP

max2010maxim [7]

Hello!

So, this is quite the complex question, and here are the following steps:

What is the quotient of $\frac{6-3\sqrt[3]{6}}{\sqrt[3]{9}}$ ?

$\frac{(6 - 3\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}$ (rationalize the denominator)

$\frac{3(2 -\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}$ (factor 3 from the expression)

$\frac{(2-\sqrt[3]{6})\sqrt[3]{9^{2}}}{3}$ (reduce the fraction with 3)

$\frac{2\sqrt[3]{9^{2}}-\sqrt[3]{9^{2}}}{3}$ (distributive property)

$\frac{2\sqrt[3]{81}-\sqrt[3]{486}}{3}$ (simplify 3 · 9²)

$\frac{6\sqrt[3]{3}-3\sqrt[3]{18}}{3}$ (simplify the radical)

$\frac{3(2\sqrt[3]{3}-3\sqrt[3]{18})}{3}$ (factor 3 from the expression)

$2\sqrt[3]{3}-\sqrt[3]{18}$ (reduce the fraction)

The answer, is simply, choice A, $2\sqrt[3]{3} -\sqrt[3]{18}$ ≈ 0.263758.

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3 years ago

2 taps A and B can fill a swimming pool in 3 hours. If turned on alone, it takes tap A 5 hours less than tap B to fill the same

vazorg [7]

Answer:

Tap A will take 2 hours and tap B will take 6 hours to fill the tank when turned on alone.

Step-by-step explanation:

Let tap B fills the pool alone in the time = x hours

So in one hour part of pool will be filled = $\frac{1}{x}$

Another tap A when turned on, it takes time to fill the pool = x-5 hours

So in one hour part of the same pool filled = $\frac{1}{x-5}$

Now both the taps A and B are turned on then time taken to fill the pool = 3 hours.

Part of the pool filled in one hour by both the taps = $\frac{1}{3}$

Now we form an equation

Part of pool filled in one hour by tap A + Part of pool filled in one hour by tap B = Part of pool filled in one hour by both the taps when turned on

$\frac{1}{(x-5)}+\frac{1}{x}=\frac{1}{3}$

$\frac{x-5+1}{x(x-5)}=\frac{1}{3}$

$\frac{x-4}{x(x-5)}=\frac{1}{3}$

3(x - 4) = x(x - 5)

x² -5x = 3x - 12

x² - 8x + 12 = 0

x² - 6x - 2x + 12 = 0

x(x - 6) - 2(x - 6) = 0

(x -2)(x - 6) = 0

x = 2, 6 hours

We will take higher value of x as x = 6 hours for tap B.

Time taken by tap A = 6 - 4 = 2 hours.

Therefore, Tap A will take 2 hours and tap B will take 6 hours to fill the tank when turned on alone.

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3 years ago

PLEASE HELP ME!!!! WORTH 10 POINTS!!!

Sidana [21]

slope
m = (1 - 1)/(0 + 2) = 0

(x1, y1) as (-2, 1)

point-slope form
y - y1 = m(x - x1)
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3 years ago