3 years ago

Sin^2x/1+cosx= with a interval 0

Mathematics

1 answer:

Art [367]3 years ago

6 0

On the interval 0 :

sin(x) = 0
sin^2(x) = 0

cos(x) = 1
1 + cos(x) = 2

sin^2(x) / [ 1 + cos(x) ] = <em>0 .</em>

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(2a² – b) (за² + 3b)

nataly862011 [7]

Answer:

a^4+3a^2-3b^2

Step-by-step explanation:

$\hookrightarrow \: (2 {a}^{2} - b)(3 {a}^{2} + 3b) \\ \hookrightarrow \:2 {a}^{2} (3 {a}^{2} + 3b) - b(3 {a}^{2} + 3b) \\ \hookrightarrow \:6 {a}^{4} + 6 {a}^{2} b - 3 {a}^{2} b - 3 {b}^{2} \\ \hookrightarrow \:{a}^{4} + 3 {a}^{2} b- 3 {b}^{2}$

6 0

2 years ago

Prove the divisibility of the following numbers:

ratelena [41]

Make use of prime factorizations:

$16^5+2^{15}=(2^4)^5+2^{15}=2^{20}+2^{15}$

Both terms have a common factor of $2^{15}$ :

$16^5+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33$

- - -

The second one is not true! We can write

$15^7+5^{13}=(3\cdot5)^7+5^{13}=3^7\cdot5^7+5^{13}$

Both terms have a common factor of $5^7$ :

$15^7+5^{13}=5^7\left(3^7+5^6\right)$

Since $30=5\cdot6$ , and $5\mid5^7$ , we'd still have to show that $5^6(3^7+5^6)$ is a multiple of 6. This is impossible, because $6=3\cdot2$ and there is no multiple of 2 that can be factored out.