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4 years ago

7

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

2 answers:

laiz [17]3 years ago

8 0

Answer:

The answer is BC < AC and BC < AB

So D and C are the answers

siniylev [52]3 years ago

4 0

The answer is BC < AC and BC < AB

I just did the test this right right.
hope I helped.

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Consider the tangent line to the curve y = 6 sin(x) at the point (π/6, 3). (a) Find a unit vector that is parallel to the tangen

mojhsa [17]

Answer:

$y=\frac{\pi }{6}+5.196(x-3)$

Step-by-step explanation:

We have given the equation y = 6 sin (x)

On differentiating both side $\frac{dy}{dx}=m=6cosx$

As it passes through the point $(\frac{\pi }{6},3)$

So $\frac{dy}{dx}=6cos\frac{\pi }{6}=5.196$

Now the unit vector is parallel to the tangent so m will be 5.196

This passes through the point $(\frac{\pi }{6},3)$

So unit vector will be $y-\frac{\pi }{6}=5.196(x-3)$

$y=\frac{\pi }{6}+5.196(x-3)$

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3 years ago

Write with fractional exponents. 6^5 square root x^2 y

sertanlavr [38]

Answer:

6xy^2/5 what im assuming is right

Step-by-step explanation:

4 0

2 years ago

What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

telo118 [61]

Answer:

$f(x)=-\frac{1}{8}(x-3)^2+3$

Step-by-step explanation:

We want to find the equation of the parabola with a focus of $(3,1)$ and directrix $y=5$ .

Considering the directrix, the quadratic graph must open downwards.

The equation of this parabola is given by the formula,

$(x-h)^2=4p(y-k)$ , where $(h,k)$ is the vertex of the parabola.

The axis of this parabola meets the directrix at $(3,5)$ .

Since the vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix,

$h=\frac{3+3}{2}=3$ and $k=\frac{5+1}{2}=3$ .

The equation of the parabola now becomes,

$(x-3)^2=4p(y-3)$ .

Also $|p|$ is the distance between the vertex and the directrix.

$|p|=2$

This implies that $p=-2\:or\:2$ .

Since the parabola turns downwards,

$p=-2$ .

Our equation now becomes,

$(x-3)^2=4(-2)(y-3)$ .

$(x-3)^2=-8(y-3)$ .

We make y the subject to get,

$y=-\frac{1}{8}(x-3)^2+3)$ .

This is the same as

$f(x)=-\frac{1}{8}(x-3)^2+3)$ .

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3 years ago

What is the value of pi to 10 decimal places

I am Lyosha [343]

Answer:

3. 415926535

Step-by-step explanation:

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3 years ago

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Hoochie [10]

Answer:

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