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sladkih [1.3K]
3 years ago
7

PLS HELP Simply the expression by combining like terms. 6x+9+2x+4=

Mathematics
1 answer:
sukhopar [10]3 years ago
4 0

Answer:

Combine like terms: 18x + 15x = 33x

(45 + 33x + 6x2)

Step-by-step explanation:

You might be interested in
A) Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)
Gelneren [198K]

For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

A)

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by <em>x</em> ²,

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}

Then each remaining rational term converges to 0 as <em>x</em> gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than <em>x</em> - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}

More precisely,

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0

D) Looks like this limit should read

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}

which is just another case of (A) and (B); the limit would be

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1

That is,

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = \lim_{t\to\infty}\frac{\dfrac1{t^{3/2}}+1}{\dfrac3t-1} = \dfrac1{-1} = -1

However, in case you meant something else, such as

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}

then the limit would be different:

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0

since the degree of the denominator is larger.

One important detail glossed over here is that

\sqrt{t^2} = |t|

for all real <em>t</em>. But since <em>t</em> is approaching *positive* infinity, we have <em>t</em> > 0, for which |<em>t</em> | = <em>t</em>.

E) Similar to (D) - bear in mind this has the same ambiguity I mentioned above, but in this case the limit's value is unaffected -

\displaystyle \lim_{x\to\infty} \frac{x^4}{\sqrt{x^8+9}} = \lim_{x\to\infty}\frac{x^4}{\sqrt{x^8}\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac{x^4}{x^4\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac1{\sqrt{1+\dfrac9{x^8}}} = \boxed{1}

Again,

\sqrt{x^8} = |x^4|

but <em>x</em> ⁴ is non-negative for real <em>x</em>.

F) Also somewhat ambiguous:

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+5x^2}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{x\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{\dfrac1x+5}}{3-\dfrac1x} = \dfrac{\sqrt5}3

or

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x}+5x^2}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{\sqrt x}+5x}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{x^{3/2}}+5}{3-\dfrac1x} = \frac53\lim_{x\to\infty}x = \infty

G) For a regular polynomial (unless you left out a denominator), the leading term determines the end behavior. In other words, for large <em>x</em>, <em>x</em> ⁴ is much larger than <em>x</em> ², so effectively

\displaystyle \lim_{x\to\infty}(x^4-2x) = \lim_{x\to\infty}x^4 = \boxed{\infty}

6 0
3 years ago
The department manager recorded the number of sick days taken last year by each employee, as shown in the table. What is the mea
musickatia [10]

Answer:

4.42857142857 but since we are rounding to the nearest hundredth it would be 4.43

Step-by-step explanation:

You would add all the numbers together (7+4+2+6+4+5+3=31) then you would divide by the number or numbers there is (31 divided by 7= 4.42857142857) then round to the nearest hundredth (4.43000000000 = 4.3)

5 0
3 years ago
PLZ HELP FAST multiple choice<br>touching spirit bear​
Juli2301 [7.4K]
Look this up
On quizlet it will give you the answer
5 0
3 years ago
When baking a cake, you have a choice of the following pans: a round cake pan that is 2 inches deep and has a 7 inch diameter a
ruslelena [56]
Round pan volume is:
3.14•r^2•h
D=7 so r=3.5 in
3.14• (3.5^2)•2 = 76.97 in^3

Rec. pan vol. is :
9•6•2= 108 in^3

Rec. Pan is larger because 108 in^3 is > 76.97 in^3 :) .

The icing that will be needed to frost the round cake pan is:
We need to find the surface area:
S.A= 3.14r^2 + 2 • 3.14•r • h .... 3.14 is the value of PI
So, S.A= 3.14• 3.5^2 + 6.28• 3.5• 2= 82.47 in^2 the icing that'll be needed to frost the round cake pan.

Icing that will be needed for the rec. cake pan is:

2•9•2=36 in^2
6•9•2= 108in^2
6•2= 12 in^2
S.A= 156 in^2 the icing needed to frost the rec. cake pan .... the S.A of all sides except the bottom one :).

Good luck ;-)

8 0
3 years ago
What Is The Meters Per Second ?
Assoli18 [71]
The answer might be 258 but idk
8 0
4 years ago
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