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3 years ago

Least to greatest 1.3,1 1/3,1.34

Mathematics

2 answers:

Valentin [98]3 years ago

6 0

Step-by-step explanation:

1.31, 1 1/3, 1.34

because 1 and one third is basically 1.3333333333 forever so its greater than 1.31 and 1.34 is greater than both 1.31 and 1 1/3

ollegr [7]3 years ago

6 0

Answer:

1.3 < 1 1/3 < 1.34

Step-by-step explanation:

1 1/3 = 1.333..., so you can order the decimal numbers from that.

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Yolanda wanted to buy a total of 6 pounds

Maslowich

Answer:

No she didn't

Step-by-step explanation:

Add the total cost of the food and divide by 6. You'll find that she is short

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3 years ago

A bag contains pennies, nickels, dimes, and quarters. There are 50 coins in all. Of the coins, 12% are pennies and 32% are

kenny6666 [7]

Answer:6.46$

Step-by-step explanation:

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3 years ago

What is an equation of the line that is perpendicular to -x+2y=4 and passes through the point (-2, 1)?

KATRIN_1 [288]

First: Slope of -x+2y=42y=x+4y=x/2+2So, m=1/2.
Second: Slope of the perpendicular line: mp=-2.
Third: Find the line with slope -2 and passes through the point (−2, 1)y-y1=m(x-x1)y-1=-2(x+2)y=-2x-4+1y=-2x-3

Read more on Brainly.com - brainly.com/question/7942650#readmore

5 0

3 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

2 years ago

Triangle ABC has coordinates A(-2, -3), B(1, 1), and C(2, -1). If the triangle is translated 4 units up, what are the coordinate

DochEvi [55]

Answer:

The coordinates of point A' will be: A'(-2, 0)

Step-by-step explanation:

We know that a translation is basically a transformation that occurs when an object is moved from one place to another place.

Translation does not change the size, orientation, or shape of the original object.

Translating a point P(x, y) 4 units up means we need to add to use P'(x, y+4), that is adding 4 units to the y-coordinate of the point P'(x, y).

Now we have to translate the triangle 4 units up. So, use the formula

P(x, y) → P'(x, y + 3)

Thus, the coordinates of point A' will be:

A(-2, -3) → A'(-2, -3+3) → A'(-2, 0)

Therefore, the coordinates of point A' will be: A'(-2, 0)

8 0

3 years ago