Can someone help me with this in the next couple minutes

On a piece of paper, use a protractor to construct △ABC with m∠A = 30° , m∠B = 60° , and m∠C = 90° .

riadik2000 [5.3K]

Here, we know largest side is always opposite to the largest angle.
As largest angle is C with 90 degree, it's opposite side which is AB must be longest here!! & AC would be second largest & BC would be smallest.

In short, Your Answer would be Option B & D

Hope this helps!

7 0

3 years ago

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What is an equation of the line that passes through the points (6,8) and (-3,2)?

Orlov [11]

Answer:

y=2/3x+4

Step-by-step explanation:

4 0

3 years ago

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Athena priced each of her art pieces at the same price and displayed them for sale in two different art shows. She made $495.80

pshichka [43]

In the question "Athena priced each of her art pieces at the same price and displayed them for sale in two different art shows. She made $495.80 from the first art show and $371.85 from the second show. If Athena sold 7 pieces of art, what price did each piece sell for?" To get the price that each piece of art sells for: Total sales made from selling 7 pieces of art = $495.80 + $371.85 = $867.65 The price for each piece of art = $867.65 / 7 = $123.95 Thus each piece of art sells for $123.95 and hence the correct answer.

4 0

4 years ago

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Please help math questions put number with answers please

natita [175]

Answer:

See below for answers

Step-by-step explanation:

Problem 1

$\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}$

Therefore, the second option is correct.

Problem 2

$cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)$

$0=sinx\\\\x=\pi n\\\\x=0,\pi$

$0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}$

Therefore, the solution set is $x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}$ , making the second option correct.

Problem 3

$2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0$

$cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ$

$2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ$