All categories

2 years ago

Is 3x+4y a polynomial. is it is a polynomial, state the degree of the polynomial.

Mathematics

2 answers:

Marina86 [1]2 years ago

6 0

Answer: 3x+4y is a binomial. The binomial is to the first degree. I believe.

Svetlanka [38]2 years ago

5 0

Yes, this is a polynomial. It is an expression containing variables raised to a real number.

In this case, the variables are being raised to the power of one.

The degree of this polynomial, thus, is one.

You might be interested in

Which is the best estimate for the width of your index finger

fgiga [73]

I'd say centimeters, if that's what you're looking for. Maybe 1-2 cm.

4 0

3 years ago

Read 2 more answers

Can u help??????????????????????????????????????//

maw [93]

Answer:

Okay!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The answer is 14

Step-by-step explanation:

Folow order of operations and when doing multipactiona or subtraction or division and addition, make sure to go from left to right

5 0

2 years ago

Read 2 more answers

What is the value of x?

AfilCa [17]

Answer:

Step-by-step explanation:

7 0

2 years ago

Perform the following steps to convert 8 years to days

Troyanec [42]

Answer:

8 years is around 2,920 days

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago