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3 years ago

Equivalent Expressions

Mathematics

1 answer:

Westkost [7]3 years ago

5 0

Tell me what grade you in and maybe I.can message u on here if your in sixth grade or lower I can help

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Write 1/10 in hundredth,fraction,decimal

Ipatiy [6.2K]

0.10
1/10

Hope I helped!

Let me know if you need anything else!

<span>~ Zoe</span>

4 0

3 years ago

Evaluate the integral of the quotient of the cosine of x and the square root of the quantity 1 plus sine x, dx.

VMariaS [17]

Answer:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c.

Step-by-step explanation:

In order to solve this question, it is important to notice that the derivative of the expression (1 + sin(x)) is present in the numerator, which is cos(x). This means that the question can be solved using the u-substitution method.

Let u = 1 + sin(x).

This means du/dx = cos(x). This implies dx = du/cos(x).

Substitute u = 1 + sin(x) and dx = du/cos(x) in the integral.

∫((cos(x)*dx)/(√(1+sin(x)))) = ∫((cos(x)*du)/(cos(x)*√(u))) = ∫((du)/(√(u)))

= ∫(u^(-1/2) * du). Integrating:

(u^(-1/2+1))/(-1/2+1) + c = (u^(1/2))/(1/2) + c = 2u^(1/2) + c = 2√u + c.

Put u = 1 + sin(x). Therefore, 2√(1 + sin(x)) + c. Therefore:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c!!!

4 0

3 years ago

Read 2 more answers

A ferritin test is a popular test to measure a person’s current iron stores. in women, ferritin is normally distributed with a m

MArishka [77]

Given:
μ = 89 ng/ml, population mean
σ = 23 ng/ml, population standard deviation

Random variable, x = 100
To test P(x < 100), calculate the z-score.
z = (x-μ)/σ = (100 - 89)/23 = 0.4783

From standard tables, obtain
P(x < 100) = 0.6838 = 68.4%

Answer: 68.4%

7 0

3 years ago

What eigen value for this matix <br> (1 -2)<br> (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$

5 0

3 years ago

Read 2 more answers

Ray needs help creating the second part of the coaster. Create a unique parabola in the pattern f(x) = (x − a)(x − b). Describe

MatroZZZ [7]

Answer:

1) The parabola equation faces up

2) The y-intercept = a·b

3) The zeros are x = a and x = b

Step-by-step explanation:

1) The given information are;

f(x) = (x - a)·(x - b) which gives;

f(x) = x² - (b+a)·x + a·b

For an equation of the form a·x² - b·x + c, if a is positive, then the parabola faces up, therefore, the parabola equation, x² - (b+a)·x + ab where a is equivalent to +1 faces up

2) The y-intercept is given at x = 0, which gives;

f(0) = 0² - (b+a)·0 + a·b

The y-intercept = a·b

3) From f(x) = (x - a)·(x - b), when f(x) = 0, we have either;

(x - a) = 0 or (x - b) = 0

Therefore;

The zeros are x = a and x = b

4 0

3 years ago