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3 years ago

Someone help me please!!

Mathematics

2 answers:

Bess [88]3 years ago

8 0

It’s 20 Bc I just googled it and did the test and got a A

vfiekz [6]3 years ago

3 0

1 - 2-n
2- 9xn-8
3- n divided by 4
4- 8-4xn
5- 14+n
6- 7xn-28
7-n divided by 9
8- 10-n
9- nx16
10- n divided by 10
11- 21+n
12- n+15
13- 3xn-1
14- 18-6xn
15- n divided by 12
16- 6+ 6xn
17- 5xn-11
18- n divided by 5
19- nx7
20- 6-nx8

I am not 100% sure this is what it means by writing the given statement in symbols but if it does mean this then these are the answers:)

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Find the value of x if the image below is a kite.

AveGali [126]

X = 12 because x+3 is equal to 15

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3 years ago

What is the answer to this?

Brut [27]

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Step-by-step explanation:

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3 years ago

F(x)=-5x-4 and g(x)= -3x-2, find (f-g)(x)

Mnenie [13.5K]

Answer:

Step-by-step explanation:

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3 years ago

The surveyor sets up the instrument to determine the height at the highest point on the hill. She determines that she is standin

Ne4ueva [31]

Answer:

$(a)\ h = d * \tan(\theta)$

$(b)\ h = 30.57m$

Step-by-step explanation:

Given

$\theta = 17^o$

$d =100m$

$h = ??$

See attachment for illustration

Solving (a): Trigonometry ratio to calculate h.

From the attachment, we have:

$\tan(\theta) = \frac{h}{d}$

Multiply both sides by d

$d * \tan(\theta) = \frac{h}{d} * d$

$d * \tan(\theta) = h$

Rewrite as:

$h = d * \tan(\theta)$

Solving (b): The value of h

We have:

$\theta = 17^o$

$d =100m$

So:

$h = 100m * \tan(17^o)$

$h = 100m * 0.3057$

$h = 30.57m$

8 0

3 years ago

Given: cosθ= -4/5 , sin x = -12/13 , θ is in the third quadrant, and x is in the fourth quadrant; evaluate tan 1/2 θ

mrs_skeptik [129]

Answer:

Step-by-step explanation:

If you're looking for what the half angle of the tangent of theta is, I'm a bit confused as to why you think the angle in the 4th quadrant, x, is relevant. But maybe you don't know it isn't and it's a "trick" to throw you off. Hmm...

Anyways, the half angle identity for tangent is

$tan(\frac{\theta}{2})=\frac{sin\theta}{1+cos\theta}$

There are actually 3 identities for the tangent of a half angle, but this one works just as well as either of the others do, so I'm going with this one.

If theta is in QIII, the value of -4 goes along the x axis and the hypotenuse is 5. That makes the missing side, by Pythagorean's Theorem, -3. Filling in our formula:

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{1+(-\frac{4}{5}) }$ which simplifies a bit to

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{\frac{5}{5} -\frac{4}{5} }$ and a bit more to

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{\frac{1}{5} }$

Bring up the lower fraction and flip it to divide to get

$tan(\frac{\theta}{2})=-\frac{3}{5}*\frac{5}{1}$ which of course simplifies to

-3. Choice A.

6 0

4 years ago