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3 years ago

Graph: y=-4sin(2x)+1 where the x axis starts at 30

Mathematics

1 answer:

gregori [183]3 years ago

8 0

Answer:

We want to graph:

y = -4*sin(2*x) + 1

Such that the x-axis starts at x = 30

This is trivial to do if we use a program to graph or if we graph by hand, here we just need to draw the x-axis such that it crosses the y-axis at the point (30, 0)

Then let's graph the equation in that axis, we will get an image like the one you can see below.

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The graph represents a function related to a train's movement over time.

siniylev [52]

Answer:

Where is the question?

Step-by-step explanation:

8 0

3 years ago

In trapezoid $ABCD$, $\overline{AB}$ is parallel to $\overline{CD}$, $AB = 7$ units, and $CD = 10$ units. Segment $EF$ is drawn

Mekhanik [1.2K]

Answer:

EF = 58/7

Step-by-step explanation:

If you extend both BC and AD to meet at P you get 3 similar triangles PAB, PEF, and PCD. Let's call BF = 3x and FC = 4x. Therefore the following proportions can be expressed:

BP/7 = (BP+7x)/10

10BP = 7*(BP+7x)

10BP = 7BP+49x

3BP = 49x

BP = (49/3)x

BP/7 = (BP+3x)/EF

EF = (BP+3x)*7/BP

EF = ((49/3)x + 3x)*7/((49/3)x)

EF = ((58/3)x*7*3)/(49x)

EF = 58/7

3 0

2 years ago

Need help with this

Eva8 [605]

Answer:

the answer is #4

Step-by-step explanation:

<h3>first you multiply the 3&4 witch is 12 then u keep the y and u add the 2&5 then u see that it is #4</h3>

4 0

3 years ago

If 37% of high school students said that the excercise regularly, find the

katrin [286]

Answer: 0.6934%

Step-by-step explanation:

Probability is the branch of mathematics that is concerned with the numerical descriptions of how likely an event will occur or how likely a proposition is true

If 37% of high school students said that the exercise regularly, the

probability that 5 students who are randomly selected will say thay they exercise regularly will be:

= (37/100)^5=0.006934

= 0.6934%

6 0

2 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago

Graph: y=-4sin(2x)+1 where the x axis starts at 30​

Graph: y=-4sin(2x)+1 where the x axis starts at 30