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3 years ago

WHAT IS 3/5 OF 734285.56=

Mathematics

1 answer:

shepuryov [24]3 years ago

7 0

Answer:

i did with a calculator to make sure

the answer is : <u>440571.336</u>

Step-by-step explanation:

hope it helps good luck in whatever math thing u r doing

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Find m∠U.<br> Write your answer as an integer or as a decimal rounded to the nearest tenth.

Ilia_Sergeevich [38]

Answer:

∠U = 56.4

Step-by-step explanation:

We can use trigonometric functions to solve this

Here we are given the opposite side of ∠U as well as the adjacent side.

When dealing with the adjacent and opposite side we use sin

Sin = Opposite side / Adjacent side

Opp = 5 and adj = 6

Sin(U) = 5/6

* take the inverse sine of both sides *

arc(sin)(u) = u

arcsin(5/6) = 56.4 *( rounded to the nearest tenth )

∠U = 56.4

5 0

2 years ago

Read 2 more answers

In which quadrant or on which axis do each of the points (-2,4) , (3,-1) ,(-1,0), (1,2) and

exis [7]

Answer:

Step-by-step explanation:

cartesian plane :

quadrant 1 : both x and y are positive

quadrant 2 : x is negative, y is positive

quadrant 3 : both x and y are negative

quadrant 4 : x is positive and y is negative

(-2,4)......we have x is negative and y is positive...so we are in quadrant 2

(3,-1)....we have x is positive and y is negative...in quadrant 4

(-1,0) ...if u have a 0 in ur points, they do not lie in any quadrant because they are located on an axis....this is located on axis x

(1,2)....both are positive....located in quadrant 1

(-3,-5)...both are negative.....quadrant 3

5 0

2 years ago

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago

Answer asap thank you!??!!??!

Semenov [28]

The answer to the problem is A

7 0

2 years ago

Find the surface area of each figure. Round answers to two decimal places. Use 3.14 for pi

ELEN [110]

Answer:

the top wpuld be 73.14 and i cant see the numbers at the bottom

Step-by-step explanation:

8 0

2 years ago