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2 years ago

5

Someone please help with this,I’m no expert at doing annotated bibliography I will give Brainliest to whoever helps me out(:

1 answer:

Nitella [24]2 years ago

5 0

An annotated bibliography includes the citation and a brief description of the source.

The citation can be made by plugging in information in an online citation website.

The description is 2 to 3 sentences about what type of source (primary or secondary) and how you used it (for the images, background info, etc).

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A triangle has side lengths of 5a +3 inches and 2a +3 inches. If the perimeter of the triangle is 3a + 12 inches, which expressi

JulsSmile [24]

Answer:

A. 2a+6

Step-by-step explanation:

I did the test and this is what I got.

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2 years ago

PLSSSSS HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Leni [432]

B is the answer, every other answer can be simplified

8 0

2 years ago

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Will mark brainliest

Lisa [10]

Answer:

1.

31, 37, 43, 49, 55

2.

for nth number the value is 31+6(n-1)

3.

24th number, 169-31=138 and 138/6=23

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2 years ago

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A wallet that holds $25 contains the same number of $1 bills, $5 bills, and quarters. How many of each type of bill or coin does

Anastasy [175]

<span>The wallet contains 4 $1 bills, 4 $5 bills, and 4 quarters.

$1 </span>× 4 = $4
$5 × 4 = $20
$0.25 × 4 = $1
$4 + $20 = $24, $24 + $1 = $25.

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2 years ago

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Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

4 0

2 years ago