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3 years ago

HELP QUICK PLEASE!!!!

Mathematics

1 answer:

Simora [160]3 years ago

7 0

Answer:

Step-by-step explanation:

Given the expression;

$3sin^2x + 4sinx - 4 = 0$

Let P = sinx

The expression becomes;

3P²+4P - 4 = 0

Factorize

3P²+6P-2P - 4 = 0

3P(P+2)-2(P+2) = 0

3P-2 = 0 and P+2 = 0

P = 2/3 and -2

When P = 2/3

sinx = 2/3

x = arcsin 2/3

x = arcsin 0.6667

x = 41.8 degrees

Also if P = -2

sinx = -2

x = arcsin (-2)

x will not exist in this case

To get other values of x

sin is positive in the second quadrant

x = 180 - 41.8

x = 138.2°

x = 360+41.8

x = 401.8°

<em>Hence the values of x within the interval are 41.8°, 138.2° and 401.8°</em>

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Which of the following is an element in the sample space for first tossing a

babunello [35]

Answer:

Step-by-step explanation:

Sample space = {H1,H2, H3, H4, H5, H6 , T1, T2, T3, T4, T5, T6}

So, (H, 6) is an element in the sample space

3 0

3 years ago

Eight cookies and 3 muffins cost $13Four cookies and six muffins cost $11.How much does a cookie cost? How much does a muffin co

NeTakaya

Answer:

the cost of each cookie is $\$1.25$

the cost of each muffin is $\$1.00$

Step-by-step explanation:

Let

x-----> the cost of each cookie

y----> the cost of each muffin

we know that

$8x+3y=13$ -----> equation A

$4x+6y=11$ -----> equation B

Using a graphing tool

The solution of the system of equations is the intersection point both graphs

The intersection point is $(1.25,1)$

see the attached figure

therefore

the cost of each cookie is $\$1.25$

the cost of each muffin is $\$1.00$

5 0

4 years ago

Who wants to be friends???

Sindrei [870]

Answer:

Are u christian?

Step-by-step explanation:

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3 years ago

In the expansion of ( x^3 - 2/x^2 ) ^10 , find the coefficient of 1/x^5

Lilit [14]

Answer:

240

Step-by-step explanation:

We need to find the coeffeicent of the binomial expansion of

$( {x}^{3} - 2 {x}^{ - 2} ) {}^{10}$

Note that

$- 2 {x}^{ - 2} = - \frac{2}{ {x}^{2} }$

The binomial theorem states that

$(x + y) {}^{n} = x {}^{n} y {}^{0} + \binom{n}{1} x {}^{n - 1} y + \binom{n}{2} x {}^{n - 2} y {}^{2} ....... + x {}^{0} y {}^{n} ( \binom{n}{n} )$

Using this, we let expand our series

$( {x}^{3} - 2 {x}^{ - 2} ) {}^{10} = x {}^{30} + \binom{10}{1} ( {x}^{27} 2 {x}^{ - 2} ) + \binom{10}{2} {x}^{24} 2x {}^{ - 4} + \binom{10}{3} {x}^{21} 2x {}^{ - 6} + \binom{10}{4} {x}^{18} 2x { }^{ - 8} + \binom{10}{5} x {}^{15} 2x {}^{ - 10} + \binom{10}{6} x {}^{12}2 x {}^{ - 12} + \binom{10}{7} x {}^{ 9} 2x {}^{ - 14} + \binom{10}{8} x {}^{ 6} 2x {}^{ - 16} + \binom{10}{9} ( {x}^{3} )2x {}^{ - 18} + 2x {}^{ - 20}$