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Ksivusya [100]
3 years ago
7

Does anybody know how to do that

Mathematics
2 answers:
viva [34]3 years ago
6 0
There is no question attached
irinina [24]3 years ago
5 0

Answer:

how to do what? there is no question attached

Step-by-step explanation:

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Can you help me? Please explain how you got it!
julia-pushkina [17]

Answer:

B.

Step-by-step explanation:

First we must find the x and y intercepts for y = -2x - 4. The y-intercept is -4, since it is b in y = mx + b, but you can also find it by plugging in zero for x:

y = -2(0) - 4 = 0 -4=-4

The x-intercept is found by plugging in zero for y:

0=-2x-4

0+4=-2x-4+4

4=-2x

\frac{4}{-2}=\frac{-2x}{-2}

-2=x

The graph with an x-intercept of -2 and a y-intercept of -4 is C. Also, less than inequalities have the shading underneath the line, so we definitely know C is the answer.

4 0
3 years ago
What are the dimensions of the following matrix [5 3]?
nexus9112 [7]
The answer is none of these
7 0
3 years ago
How do you divide 9,962 and 41 how do you divide 909900 + 62 + 41
7nadin3 [17]
242.975609756 I don't know about the last one though.
5 0
3 years ago
What number times 0.1 = 0.02
guajiro [1.7K]

Answer: 0.2

Step-by-step explanation:

6 0
3 years ago
For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?
Bond [772]
Given that r=\theta, then r'=1

The slope of a tangent line in the polar coordinate is given by:

m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}

Thus, we have:

m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}



Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

\sin\theta+\theta\cos\theta=0 \\  \\ \theta\cos\theta=-\sin\theta \\  \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488



Part B:

For vertical tangent lines, \frac{1}{m} =0

Thus, we have:

\cos\theta-\theta\sin\theta=0 \\  \\ \Rightarrow\theta\sin\theta=\cos\theta \\  \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>
3 0
3 years ago
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