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4 years ago

In don't understand how to work this problem out

Mathematics

1 answer:

andrezito [222]4 years ago

4 0

$x^2+8x-11=\\
x^2+8x+16-27=\\
(x+4)^2-27=\\
(x+4-\sqrt{27})(x+4+\sqrt27)=\\
(x+4-3\sqrt{3})(x+4+3\sqrt3)$

$x^2+6x-11=\\
x^2+6x+9-20=\\
(x+3)^2-20=\\
(x+3-\sqrt{20})(x+3+\sqrt{20})=\\
(x+3-2\sqrt5)(x+3+2\sqrt5)$

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VikaD [51]

2.30 + 0.40(x - 1) < = 5.10
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4 years ago

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3 years ago

Convert 8% to a fraction/mixed number and a decimal. Select all that apply.

mr Goodwill [35]

Answer:

hope this helps :)

Step-by-step explanation:

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3 years ago

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lord [1]

Answer:

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Step-by-step explanation:

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3 years ago

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Question 20 not yet answered marked out of 1.00 not flaggedflag question question text luna created a trash can in the shape of

Alex Ar [27]

Consider the triangle with sides a=1.6, b=2.3 and c=1.2 feet. You can use cosine theorem twice to find two unknown angles:

$b^2=a^2+c^2-2ac\cos \angle B,\\ (2.3)^2=(1.6)^2+(1.2)^2-2\cdot 1.6\cdot 1.2\cos \angle B,\\ 5.29=2.56+1.44-3.84\cos \angle B,\\ 3.84\cos \angle B=4-5.29=-1.29,\\\cos \angle B=-\dfrac{1.29}{3.84} =-0.336$

Then you can conclude that ∠B is obtuse (because $\cos \angle B$ ) and m∠B=110°.

$c^2=a^2+b^2-2ab\cos \angle C,\\ (1.2)^2=(1.6)^2+(2.3)^2-2\cdot 1.6\cdot 2.3\cos \angle C,\\ 1.44=2.56+5.29-7.36\cos \angle C,\\ 7.36\cos \angle C=7.85-1.44=6.41,\\\cos \angle C=\dfrac{6.41}{7.36} =0.871$ .

Then you can conclude that ∠C is acute (because $\cos \angle C>0$ ) and m∠C=29°.

3. m∠A+m∠B+m∠C=180°, then m∠A=180°-110°-29°=41°.

Answer: m∠A=41° (∠A is opposite to the side a=1.6), m∠B=110° (∠B is opposite to the side b=2.3) and m∠C=29° (∠C is opposite to the side c=1.2)

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3 years ago