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3 years ago

Choose the set of equivalent fractions that correctly uses the LCD for these fractions: 2/3 and 1/6

Mathematics

1 answer:

andreev551 [17]3 years ago

4 0

Answer:

4/6 and 1/6

Step-by-step explanation:

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Select the two binomials that are factors of this trinomial.

Stella [2.4K]

we are given

trinomial as

$x^2 -x -12$

now, we can factor it

$x^2 -x -12= x^2 -4x +3x -4*3$

$x^2 -x -12= x(x -4) +3(x -4)$

$x^2 -x -12= (x-4)(x+3)$

so, option-C and D ..............Answer

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4 years ago

Find the measure of the angle shown by the arrow(in radians)

soldi70 [24.7K]

Answer:

-7π/3 or -420 degrees

Step-by-step explanation:

So we know that the arrow has made one full rotation and that it is moving in a clockwise direction. So already we have -2π degrees.

The arrow stops at π/6 before the next π/2 rotation. Therefore we find the difference between the two.

π/2 - π/6

3π/6 - π/6 = 2π/6

π/3

Since this is still clockwise, we make this negative. So the measure of the angle shown by the arrow is -2π - π/3

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3 years ago

Analyze the diagram below and complete the instructions that follow.

umka2103 [35]

Answer:

B. 4/5

Step-by-step explanation:

Sin = Opposite/Hypotenuse

Sin∠A = 8/10 = 4/5

8 0

3 years ago

Need help I dont understand hep please

Travka [436]

The formulas
are on the picture

5 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago