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2 years ago

Please help me I have been stuck for a while

Mathematics

1 answer:

tatyana61 [14]2 years ago

4 0

Answer:

∛49.

Step-by-step explanation:

1/2 2 and √9 ( = 3) are rational.

∛49 = 3.65930571......

- the number goes on without bounds.

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Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa

Nataly_w [17]

Answer:

The first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

coefficient of $x^{2}$ in the expansion of $(2+x - x^{2})^{6}$ = (240 - 192) = 48

Step-by-step explanation:

$(2+x)^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})$

= $6_{C_{0}} \times x^{0} \times 2^{6} + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4}$ + terms involving higher powers of x

= $64 + 192 \times x^{1} + 240 \times x^{2}$ + terms involving higher powers of x

so, the first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

Again,

$(2+x - x^{2})^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times (2 + x)^{k} \times (-x^{2})^{6 - k})$

Now, by inspection,

the term $x^{2}$ comes from k =5 and k = 6

for k = 5, the coefficient of $x^{2}$ is , $(-32) \times 6$ = -192

for k = 6 , the coefficient of $x^{2}$ is, $6_{C_{2}} \times 2^{4}$ = 240

so, coefficient of $x^{2}$ in the final expression = (240 - 192) = 48

3 0

2 years ago

Ricardo is constructing a line through point P that is perpendicular to line m. He has already constructed the arc shown.

gogolik [260]

I took this test and got 100% it is #4 :)

3 0

3 years ago

Read 2 more answers

Find the angles between -pi and pi that satisfy:<br><br> (−√3) sin(v)+cos(v)=√3

MAXImum [283]

Observe that

$\sin\left(\dfrac\pi6-v\right)=\sin\dfrac\pi6\cos v-\cos\dfrac\pi6\sin v=\dfrac12\cos v-\dfrac{\sqrt3}2\sin v$

In the original equation, divide both sides by $\dfrac12$ :

$-\sqrt3\sin v+\cos v=\sqrt3\implies\dfrac12\cos v-\dfrac{\sqrt3}2\sin v=\dfrac{\sqrt3}2$

$\implies\sin\left(\dfrac\pi6-v\right)=\dfrac{\sqrt3}2$

Next,

$\sin x=\dfrac{\sqrt3}2\implies x=\dfrac\pi3+2n\pi,x=\dfrac{2\pi}3+2n\pi$

where $n$ is any integer. Then

$\sin\left(\dfrac\pi6-v\right)\implies v=-\dfrac\pi6-2n\pi,v=-\dfrac\pi2-2n\pi$

Fix $n=0$ to ensure $-\pi$ , so that

$v=-\dfrac\pi6,v=-\dfrac\pi2$

3 0

3 years ago

I think it is A but i am not sure can someone check and tell me if im wrong

nordsb [41]

Answer: it is 1.8

Step-by-step explanation: you can look up this same qustion and find it on branly <3 your right

7 0

1 year ago

The sum of two polynomials is 10a2b2 – 8a2b 6ab2 – 4ab 2. if one addend is –5a2b2 12a2b – 5, what is the other addend? 15a2b2 –

Brrunno [24]

The other polynomial addend, when the sum of two polynomials is 10a2b2 – 8a2b 6ab2 – 4ab is 15a²b²- 20a²b + 6ab² - 4ab + 7.

<h3>What is polynomial?</h3>

Polynomial equations is the expression in which the highest power of the unknown variable is n (n is real number).

The sum of two polynomials is,

$10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2$

The one polynomial addend is,

$-5a^2b^2 +12a^2b - 5$

Let suppose the other polynomial addend is f(a,b). Thus,

$(-5a^2b^2 +12a^2b - 5)+f(a,b)=(10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2)$

Isolate the second polynomial as,

$f(a,b)=(10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2)-(-5a^2b^2 +12a^2b - 5)\\f(a,b)=10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2+5a^2b^2 -12a^2b + 5$

Arrange the like terms as,

$f(a,b)=10a^2b^2+5a^2b^2 - 8a^2b -12a^2b+ 6ab^2 - 4ab + 2+5\\f(a,b)=15a^2b^2- 20a^2b + 6ab^2 - 4ab + 7$

Hence, the other polynomial addend, when the sum of two polynomials is 10a2b2 – 8a2b 6ab2 – 4ab is 15a²b²- 20a²b + 6ab² - 4ab + 7.

Learn more about polynomial here;

brainly.com/question/24380382

7 0

2 years ago