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2 years ago

How many terms are in the binomal expansion of (2x-3)^5

Mathematics

2 answers:

Anastasy [175]2 years ago

7 0

Answer:

Step-by-step explanation:

If we are raising to the 5th power we have to allow for the constant at the end of it all.

marysya [2.9K]2 years ago

6 0

Answer:

The answer is 6.

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What is the cube root for 3000??

Alona [7]

$\sqrt[3]{3000}=\sqrt[3]{1000\times3}=\sqrt[3]{1000}\times\sqrt[3]3=10\sqrt[3]3$

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3 years ago

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3 + 3(k + 3) = 6(k - 2) +9

ANTONII [103]

Step-by-step explanation:

3 + 3(k + 3) = 6(k - 2) + 9

3 + 3k + 9 = 6k - 12 + 9

12 + 12 - 9 = 6k - 3k

15 = 3k

k = 15/3

k = 5

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2 years ago

The transformation of 'f' is represented by 'g'. Write a rule for g. f(x)=\root(3)(x). Then show your check steps.

podryga [215]

Answer:

$g(x)=-2\sqrt[3]x$

$g(x) = -2f(x)$

Step-by-step explanation:

Given

$f(x) = \sqrt[3]x$

Required

Write a rule for g(x)

See attachment for grid

From the attachment, we have:

$(x_1,y_1) = (-1,2)$

$(x_2,y_2) = (1,-2)$

We can represent g(x) as:

$g(x) = n * f(x)$

So, we have:

$g(x) = n * \sqrt[3]x$

For:

$(x_1,y_1) = (-1,2)$

$2 = n * \sqrt[3]{-1}$

This gives:

$2 = n * -1$

Solve for n

$n = \frac{2}{-1}$

$n = -2$

To confirm this value of n, we make use of:

$(x_2,y_2) = (1,-2)$

So, we have:

$-2 = n * \sqrt[3]1$

This gives:

$-2 = n * 1$

Solve for n

$n = \frac{-2}{1}$

$n = -2$

Hence:

$g(x) = n * \sqrt[3]x$

$g(x)=-2\sqrt[3]x$

or:

$g(x) = -2f(x)$

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2 years ago

An arc of a circle is 3πm long, and it subtends an angle of 72° at the center of the circle. Find the radius of the circle.

Alex777 [14]

72⁰_______3π m
360⁰______15π m

C = 15π = 2π r
r = 7•5 m

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3 years ago

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Colt1911 [192]

Answer:

48 il take brainliest instead lol

Step-by-step explanation:

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3 years ago

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