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3 years ago

PLEASE HELP!!!!

Mathematics

2 answers:

den301095 [7]3 years ago

7 0

Answer:

multiplication

Step-by-step explanation:

valina [46]3 years ago

7 0

Hopes this helps:

Answer: multiplication

We can multiply two fractions with different denominators, but we cannot add and subtract them.

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If the coordinates (-2, 1), (1, 1), and (1, 4) are translated 5 units right, the new coordinates would be ______? *If the coordi

Natali5045456 [20]

I would say B . From the 5 Units

4 0

3 years ago

Read 2 more answers

a 90% confidence interva is constructed for the population mean. If a 95% confidence interval had been constructed instead (ever

zhuklara [117]

Answer: increased

Step-by-step explanation:

An x% confidence interval indicates that a person can be x% confident that true population parameter lies in it.

More level of confidence more width of the interval.

As level of Confidence interval increases width of interval increases.

Width of interval $\propto$ level of Confidence interval

So, If a 95% confidence interval had been constructed instead of 90% the width of the interval would have been<u> increased.</u>

6 0

3 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

3 0

2 years ago

I need help setting up my problem

Genrish500 [490]

The measure of angle A is 144.3 degrees and the angle to cut the molding is 54.3 degrees

<h3>How to solve for angle A?</h3>

Start by solving the acute part of angle A using the following sine function

sin(Ax) = (30 - 4)/32

Evaluate the quotient

sin(Ax) = 0.8125

Take the arc sin of both sides

Ax = 54.3

The measure of angle A is then calculated as:

A = 90 + Ax

This gives

A = 90 + 54.3

Evaluate

A = 144.3

Hence, the measure of angle A is 144.3 degrees

<h3>The angle to cut the molding</h3>

In (a), we have:

Ax = 54.3

This represents the angle where the molding would be cut

Hence, the angle to cut the molding is 54.3 degrees