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3 years ago

All graphs need to have a key, the data (imformation) and a _____?

Mathematics

1 answer:

jarptica [38.1K]3 years ago

6 0

Answer:

Step-by-step explanation:

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Is 5266 divisible by 2

Gemiola [76]

Answer:

yes it is, the answer is 2633

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4 years ago

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What adds to -288 and adds to -41?

inna [77]

-329 hood this helps

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2 years ago

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Find, correct to four decimal places, the length of the curve of intersection of the cylinder 4x2 1 y2 − 4 and the plane x 1 y 1

charle [14.2K]

<u>Answer-</u> Length of the curve of intersection is 13.5191 sq.units

<u>Solution-</u>

As the equation of the cylinder is in rectangular for, so we have to convert it into parametric form with

x = cos t, y = 2 sin t (∵ 4x² + y² = 4 ⇒ 4cos²t + 4sin²t = 4, then it will satisfy the equation)

Then, substituting these values in the plane equation to get the z parameter,

cos t + 2sin t + z = 2

⇒ z = 2 - cos t - 2sin t

∴ $\frac{dx}{dt} = -\sin t$

$\frac{dy}{dt} = 2 \cos t$

$\frac{dz}{dt} = \sin t-2cos t$

As it is a full revolution around the original cylinder is from 0 to 2π, so we have to integrate from 0 to 2π

∴ Arc length

$= \int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}$

$=\int_{0}^{2\pi}\sqrt{(-\sin t)^{2}+(2\cos t)^{2}+(\sin t-2\cos t)^{2}$

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$

Now evaluating the integral using calculator,

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$ $= 13.5191$

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3 years ago

Find the measure of ∠P

Blizzard [7]

Triangles internal angles will always equal 180 degrees.

So, with that being said we can add up the known angles and subtract them from 180 to get our missing angle

28+67 = 95
180-95 = 85
so angle P is 85 degrees

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3 years ago

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Can you help me to solve this

Ber [7]

Answer:

$\frac{4a-2}{2a+1}$

Step-by-step explanation:

Factorise the numerator and denominator

8a² - 2 ← factor out 2 from each term

= 2(4a² - 1) ← 4a² - 1 is a difference of squares

= 2(2a - 1)(2a + 1)

4a² + 4a + 1 ← is a perfect square

= (2a + 1)²

Thus

$\frac{8a^2-2}{4a^2+4a+1}$

= $\frac{2(2a-1)(2a+1)}{(2a+1)(2a+1)}$ ← cancel (2a + 1) on numerator/ denominator

= $\frac{2(2a-1)}{2a+1}$

= $\frac{4a-2}{2a+1}$

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3 years ago

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