All categories

3 years ago

What is the classification of the polynomial by the number of terms?

Mathematics

1 answer:

hichkok12 [17]3 years ago

8 0

Answer: Trinomial

Step-by-step explanation:

You might be interested in

781÷26<br>26×3=78<br>is answer 3 remainder 1 or 30 remainder 1

mamaluj [8]

Which one is the one you want, 781:-:(divison)26?

8 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

Find the coordinates of point X that lies along the directed line segment from Y(-8, 8) to T(-15, -13) and partitions the segmen

Nataly_w [17]

Answer:

C. (-13, -7)

Step-by-step explanation:

The location of a point O(x, y) that divides a line AB with location A $(x_1,y_1)$ and B $(x_2,y_2)$ in the ratio m:n is given by:

$x=\frac{m}{m+n} (x_2-x_1)+x_1\\\\y=\frac{m}{m+n} (y_2-y_1)+y_1$

Therefore the coordinates of point X That divides line segment from Y(-8, 8) to T(-15, -13) in the ratio 5:2 is:

$x=\frac{5}{5+2} (-15-(-8))+(-8)\\\\x=\frac{5}{7} (-15+8)-8=\frac{5}{7}(-7)-8=-5-8=-13 \\\\\\y=\frac{5}{5+2} (-13-8)+8\\\\y=\frac{5}{7} (-21)+8=5(-3)+8=-15+8=-7$

Therefore the coordinates of point X is at (-13, -7)

6 0

3 years ago

X+y=c x/y=p/q please find x and y

Paul [167]

Step-by-step explanation:

x=c-y. / x=py/q

y=c-x. / y=xq/p

7 0

3 years ago

Read 2 more answers

Anyone out there doing Friday school one this sucks two

Rufina [12.5K]

Answer:

thr end

Step-by-step explanation:

6 0

3 years ago