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3 years ago

Important !

Mathematics

1 answer:

Shkiper50 [21]3 years ago

6 0

2/10 is what i think it is idk if it’s correct or not

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Solve for x <br>40 = x/ -4/7<br>

Lena [83]

Answer:

$40 = \frac{x}{ - \frac{4}{7} } \\ 40 \times ( - \frac{4}{7} ) = x \\ x = - \frac{160}{7}$

3 0

2 years ago

What is the answer to this question I can’t work it out Ni=n02(yi-yo)/t2

mr_godi [17]

Changing the subject of an equation involves solving for another variable in the equation

The equation of t is $t=\frac{\sqrt{y_i-y_o}}{N}$

<h3>How to change the subject of the equation</h3>

The equation is given as:

$N=\sqrt[2]{\frac{y_i-y_o}{t^2}}$

Rewrite the equation as:

$N=\sqrt{\frac{y_i-y_o}{t^2}}$

Evaluate the square root of t^2

$N=\frac{\sqrt{y_i-y_o}}{t}$

Multiply both sides by t

$Nt=\sqrt{y_i-y_o}$

Make t the subject, in the above equation

$t=\frac{\sqrt{y_i-y_o}}{N}$

Hence, the equation of t is $t=\frac{\sqrt{y_i-y_o}}{N}$

Read more about subject of formula at:

brainly.com/question/657646

6 0

2 years ago

Find a parametric representation for the part of the cylinder y2 + z2 = 49 that lies between the planes x = 0 and x = 1. x = u y

sweet-ann [11.9K]

Answer:

The equation for z for the parametric representation is $z = 7 \sin (v)$ and the interval for u is $0\le u\le 1$ .

Step-by-step explanation:

You have the full question but due lack of spacing it looks incomplete, thus the full question with spacing is:

Find a parametric representation for the part of the cylinder $y^2+z^2 = 49$ , that lies between the places x = 0 and x = 1.

$x=u\\ y= 7 \cos(v)\\z=? \\ 0\le v\le 2\pi \\ ?\le u\le ?$

Thus the goal of the exercise is to complete the parameterization and find the equation for z and complete the interval for u

Interval for u

Since x goes from 0 to 1, and if x = u, we can write the interval as

$0\le u\le 1$

Equation for z.

Replacing the given equation for the parameterization $y = 7 \cos(v)$ on the given equation for the cylinder give us

$(7 \cos(v))^2 +z^2 = 49 \\ 49 \cos^2 (v)+z^2 = 49$

Solving for z, by moving $49 \cos^2 (v)$ to the other side

$z^2 = 49-49 \cos^2 (v)$

Factoring

$z^2 = 49(1- \cos^2 (v))$

So then we can apply Pythagorean Theorem:

$\sin^2(v)+\cos^2(v) =1$

And solving for sine from the theorem.

$\sin^2(v) = 1-\cos^2(v)$

Thus replacing on the exercise we get

$z^2 = 49\sin^2 (v)$

So we can take the square root of both sides and we get

$z = 7 \sin (v)$

4 0

3 years ago

The perimeter of a square measures 1.3 meters. What is its side length in centimeters?

ololo11 [35]

Side length in centimeters is

A) 130 cm

5 0

3 years ago

What is this equal<br> how can I solve similar trigonometric integrals like this one

Angelina_Jolie [31]

Answer:

ln|sec θ + tan θ| + C

Step-by-step explanation:

The integrals of basic trig functions are:

∫ sin θ dθ = -cos θ + C

∫ cos θ dθ = sin θ + C

∫ csc θ dθ = -ln|csc θ + cot θ| + C

∫ sec θ dθ = ln|sec θ + tan θ| + C

∫ tan θ dθ = -ln|cos θ| + C

∫ cot θ dθ = ln|sin θ| + C

The integral of sec θ can be proven by multiplying and dividing by sec θ + tan θ, then using ∫ du/u = ln|u| + C.

∫ sec θ dθ

∫ sec θ (sec θ + tan θ) / (sec θ + tan θ) dθ

∫ (sec² θ + sec θ tan θ) / (sec θ + tan θ) dθ

ln|sec θ + tan θ| + C

3 0

3 years ago