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3 years ago

Round 1289.48 to the nearest dollar please

Mathematics

2 answers:

Ilya [14]3 years ago

8 0

$1290 or you could use 1289.50

svp [43]3 years ago

5 0

Answer:

$1,289 is your answer.

Step-by-step explanation:

Since the 4 is less than 5, the 9 stays a 9.

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If the area of square 1 is 250 units squared, and the area of square 3 is 120 units squared, what is the area of square 2? Expla

ExtremeBDS [4]

Somebody else had the same question but if you want a detailed answer, then go to this link

brainly.com/question/13227863

8 0

3 years ago

B) Complete the table to show the total amount of calories Mary will burn for

nevsk [136]

For 0 minutes she can burn 0 calories

For 10 minutes she can burn 20-40 calories

For 15 mins she can burn 40-60 calories

For 30 minutes she can burn 140-295 calories

I hoped this helped!

Have a great day/night! :)

7 0

2 years ago

Match the parametric equations with the verbal descriptions of the surfaces by putting the letter of the verbal description to t

alukav5142 [94]

Answer:

1. B

2. C

3. A

4. D

Step-by-step explanation:

The parametric equations of the circular cylinder are:

$x(u,v)=a\cos v\\y(u,v)=a\sin v\\z(u,v)=u$

If the orientation of the cylinder is changed to have the height $u$ along the x-axis, the parametric equations of the cylinder match:

$3. \mathbf{r} \left( u, v \right) = u \mathbf{i} + \cos v \mathbf{j} + \sin v \mathbf{k}$

The parametric equations of the circular paraboloid are:

$x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=u^2$

Using the units vectors the parametric equations match:

$1. \mathbf{r} \left( u, v \right) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^{2} \mathbf{k}$

The parametric equations of the cone are:

$x(u,v)=au\cos v\\y(u,v)=au\sin v\\z(u,v)=u$

Using the units vectors and rotating the base of the cone from $z=0$ to $x=0$ the parametric equations match:

$2. \mathbf{r} \left( u, v \right) = u \mathbf{i} + u \cos v \mathbf{j} + u \sin v \mathbf{k}$

The equation left is the equation of a plane:

$4. \mathbf{r} \left( u, v \right) = u \mathbf{i} + v \mathbf{j} + \left( 2u - 3v \right) \mathbf{k}$

5 0

3 years ago

(cos2A−cos2B)2+(sin2A+sin2B)2=4sin2(A+B)

lbvjy [14]

Answer:

${( \cos2A - \cos2B) }^{2} + {( \sin2A + \sin2 B }^{2}\\ = ( - 2\cos2A \cos2B + { \cos {}^{2} 2A } + \cos {}^{2} 2B) + (2 \sin2A \sin2B + \sin {}^{2} 2A + \sin {}^{2} 2B)\\ \\ { = - 2( \sin2A + \sin 2 B - \cos2 A - \cos2B) }\\ \\ { = - 2( - 4 \sin( A + B) \cos(A + B) } \\ \\ = {\tt{double \: angle \: of \: sine}}\\ \\ { = 2 \times 2(2 \sin (A + B) \cos(A + B)) }\\ \\ { = 4 \sin2(A + B)}$