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4 years ago

2sin^(-2490)+tan 1410

1 answer:

7 0

2sin²(-2490)+tan 1410=2sin²(-330-6*360)+tan (330+3*360)=
=2sin²(-330)+tan (330)=2 sin²(30-1*360)+tan (-30)=
=2sin²(30)-tan (30)=2(1/2)²-√3/3=1/2-√3/3)=(3-2√3)/6

Answer: 2 sin²(-2490)+tan (1410)=(3-2√3)/6 (≈-0.07735...)

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Answer:

88 units

Step-by-step explanation:

The required perimeter is 88 units.

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3 years ago

Isosceles triangle ABC is shown below. Triangle A B C. Angles B and C are x. Sides C A and B A extend to form an exterior angle

Mnenie [13.5K]

Answer:

79°

Step-by-step explanation:

Assuming the diagram need to complete the question is the same as the one attached below,

From the diagram,

∠A+∠B+∠C = 180 (sum of the angles in a triangle is 180)..................... Equation 1

Given: ∠A = 22° (Vertically opposite angles are equal), ∠B = x, ∠C = x

Substitute these values into equation 1

22+x+x = 180

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7 0

3 years ago

Read 2 more answers

Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2

inna [77]

$(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)$

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

$6 x^{6}/3x = 2 x^{5} \\ \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\$

We repeat previous steps until we run out of numbers:
$3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2$
$\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}$

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
$(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )$

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3 years ago

Pls help rationalize

Zepler [3.9K]

Answer:

1= 0.109

2= 3.46

3 = -(minus)1.24

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers