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3 years ago

What is the domain of the function f(x)=x^2+3x+5?

Mathematics

2 answers:

Ugo [173]3 years ago

8 0

Answer:

all real numbers

Step-by-step explanation:

Kitty [74]3 years ago

4 0

Answer:

The domain of this function is all real numbers

Step-by-step explanation:

Unless there is a domain restriction for the function, the domain of a function is all real numbers. This means that all we need to do is determine whether there is a domain restriction.

The things that cause a domain restriction are: an x in the denominator of a fraction like $f(x)=\frac{1}{x}$ or a radical like $f(x)=\sqrt{1-x}$

As there are no variables in the denominator or a radical, we can conclude that the domain of this function is all real numbers.

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Use the distributive property to rewrite (5x + 3)(2) as a sum of two terms. please hurry

Anna007 [38]

Answer:

10x + 6

Step-by-step explanation:

When distributing, you need to multiply the number on its own by both terms of the binomial (the 5x + 3). 5 times 2 is 10, and then you have the x, so it is 10x. Multiply 3 times 2 and get 6. Hope this helps!

6 0

3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

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

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago

What is the answer to 6 – 4r = -3r

hodyreva [135]

Let's solve for "r" by bringing them all to one side.
-3r=6-4r
+4r +4r
r=6
So, r is equal to 6.

7 0

3 years ago

Read 2 more answers

Evaluate - (2-5)(23). A. 4 B. -4 O c. 4 O D.

Simora [160]

Answer:

OPTION A

I hope this helps you...

6 0

3 years ago

Can I get some help with number 2 please :/

alexandr1967 [171]

Hello,

So the original temperature was 7 degrees Celsius.

Then, it went down 10 degrees.

7-10= -3

So, in this case, the answer is C) -3 degrees celsius

5 0

3 years ago

Read 2 more answers