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3 years ago

Find the limit of the function by using direct substitution.

Mathematics

1 answer:

musickatia [10]3 years ago

5 0

Lim x->1 (x^2+8x-2)

substitute x =1
= 1^2 +8(1)-2
= 1 +8 -2
= 7

So the limit = 7

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7. Eleven students go to lunch. There are two circular tables in the dining hall, one can seat 7 people, the other can hold 4. I

Ipatiy [6.2K]

Answer:

239,580 ways of seating

Step-by-step explanation:

11 students will be divided into 2 groups. One group of 7 people and one group of 4 people. So first we need to find the number of ways of dividing 11 students into these 2 groups.

First group is of 7 people. We have to select 7 people out of 11. The order of selection does not matter so this is a combination problem. Selecting 7 people from 11 can be expressed as 11C7.

Formula for combination is:

$^{n}C_{r}=\frac{n!}{r!(n-r)!}$

For the given case this would be:

$^{11}C_{7}=\frac{11!}{7! \times 4!}=330$

So, there are 330 ways of selecting a group of 7 from 11 students. When these 7 students are selected the remaining 4 will go to the other group. So, we can say there are 330 ways to divide the 11 students in groups of 7 and 4. Note that if you start with group of 4 students, the answer will still the same because 11C4 is also equal to 330.

Next we have to arrange 7 students on a round table. The number of possible arrangements would be = (7 - 1)! = 6! = 720

Similarly, to arrange 4 people on a round table, the number of possible arrangements would be = (4 - 1)! = 3! = 6

Since, for each selection of the 330 groups, there are 720 + 6 possible seating arrangements, so the total number of possible seating arrangements would be:

330 ( 720 + 6) = 239,580 ways

Thus, there are 239,580 ways of seating 11 students.

7 0

3 years ago

Cindy's Gift Shoppe purchased 8-inch candles at $9.95 a dozen, less a 25 percent trade discount. If the owner wishes to sell the

Diano4ka-milaya [45]

The first thing you should do for this case is to take 35% out of 9.95
We have then
(0.35) * (9.95) = 3.4825
Then, we must add this value from the original price of a dozen
9.95+3.4825 = 13.4325
less a 25 percent trade discount
 13.4325*(1-0.25)= 10.074375
 answer
the selling price of each dozen should be $ 10.074375

6 0

3 years ago

How do you solve this? Thank you

V125BC [204]

2)

a)

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
(4x^5\cdot x^{\frac{1}{3}})+(2x^4\cdot x^{\frac{1}{3}})-(7x^3\cdot x^{\frac{1}{3}})+(3x^2\cdot x^{\frac{1}{3}})\\\\+(9x^1\cdot x^{\frac{1}{3}})-(1\cdot x^{\frac{1}{3}})
\\\\\\
4x^{5+\frac{1}{3}}+2x^{4+\frac{1}{3}}-7x^{3+\frac{1}{3}}+9x^{1+\frac{1}{3}}-x^{\frac{1}{3}}$

$\bf 4x^{\frac{16}{3}}+2x^{\frac{13}{3}}-7x^{\frac{10}{3}}+9x^{\frac{4}{3}}-x^{\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{16}}+2\sqrt[3]{x^{13}}-7\sqrt[3]{x^{10}}+9\sqrt[3]{x^4}-\sqrt[3]{x}$

b)

$\bf \cfrac{4x^5+2x^4-7x^3+3x^2+9x-1}{x^{\frac{1}{3}}}\impliedby \textit{distributing the denominator}
\\\\\\
\cfrac{4x^5}{x^{\frac{1}{3}}}+\cfrac{2x^4}{x^{\frac{1}{3}}}-\cfrac{7x^3}{x^{\frac{1}{3}}}+\cfrac{3x^2}{x^{\frac{1}{3}}}+\cfrac{9x}{x^{\frac{1}{3}}}-\cfrac{1}{x^{\frac{1}{3}}}
\\\\\\
(4x^5\cdot x^{-\frac{1}{3}})+(2x^4\cdot x^{-\frac{1}{3}})-(7x^3\cdot x^{-\frac{1}{3}})+(3x^2\cdot x^{-\frac{1}{3}})\\\\+(9x^1\cdot x^{-\frac{1}{3}})-(1\cdot x^{-\frac{1}{3}})$

$\bf 4x^{5-\frac{1}{3}}+2x^{4-\frac{1}{3}}-7x^{3-\frac{1}{3}}+9x^{1-\frac{1}{3}}-x^{-\frac{1}{3}}
\\\\\\
4x^{\frac{14}{3}}+2x^{\frac{11}{3}}-7x^{\frac{8}{3}}+9x^{\frac{2}{3}}-x^{-\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{14}}+2\sqrt[3]{x^{11}}-7\sqrt[3]{x^{8}}+9\sqrt[3]{x^{2}}-\frac{1}{\sqrt[3]{x}}$

3)

$\bf \begin{cases}
f(x)=\sqrt{x}-5x\implies &f(x)x^{\frac{1}{2}}-5x\\\\
g(x)=5x^2-2x+\sqrt[5]{x}\implies &g(x)=5x^2-2x+x^{\frac{1}{5}}
\end{cases}
\\\\\\
\textit{let's multiply the terms from f(x) by each term in g(x)}
\\\\\\
x^{\frac{1}{2}}(5x^2-2x+x^{\frac{1}{5}})\implies x^{\frac{1}{2}}5x^2-x^{\frac{1}{2}}2x+x^{\frac{1}{2}}x^{\frac{1}{5}}$

$\bf 5x^{\frac{1}{2}+2}-2x^{\frac{1}{2}+1}+x^{\frac{1}{2}+\frac{1}{5}}\implies \boxed{5x^{\frac{5}{2}}-2x^{\frac{3}{2}}+x^{\frac{7}{10}}}
\\\\\\
-5x(5x^2-2x+x^{\frac{1}{5}})\implies -5x5x^2-5x2x+5xx^{\frac{1}{5}}
\\\\\\
-25x^3+10x^2-5x^{1+\frac{1}{5}}\implies \boxed{-25x^3+10x^2-5x^{\frac{6}{5}}}$

$\bf 5\sqrt{x^5}-2\sqrt{x^3}+\sqrt[10]{x^7}-25x^3+10x^2-5\sqrt[5]{x^6}$

6 0

3 years ago

1. A boat drifted 15 m out into the water off Ocean Beach, where the shore drops off

Lunna [17]

The correct responses found using trigonometric ratios are as follows;

Brooke's cannot bring the boat back without getting water over her boots.
The plan will not be approved
The plane will not disturb
The company can not use the chimney legally
The Long company will build the bridge
The fireman can reach her.
The lift does not take the skiers up to the bar.
Eric will be able to see his Dad's plane.
The 73 m. brace

<h3>How can trigonometric ratios be used to find the right responses?</h3>

1. The depth of the water = 15 × tan(4°) = 1.049 m

The height of Brooke's boot = 1 m

Therefore;

Brooke's cannot bring the boat back without getting water over her boots.

2. The angle of elevation of the plan = arccos(99/100) ≈ 8.110°

Given that the angle of elevation the plan, 8.110°, is more than 8° limit, we find;

The plan will not be approved.

3. Height reached by the plane = 9 × tan(21°) = 3.455

Therefore;

The plane will not disturb.

4. Height of the chimney at the factory = 50 × tan(51°) = 61.745

The minimum chimney height = 65 m.

Therefore;

The company can not use the chimney legally.

6. Length of the bridge = 72 × tan(55°) ≈ 102.83

Therefore;

The Long company will build the bridge because it is longer than 100 m. but not more than 103 m.

7. Height reached by the ladder = 26 × sin(43°) ≈ 17.7

Therefore;

The fireman can reach her.

8. Height reached by the ski lift = 1000 × sin(29°) ≈ 484.8

Therefore;

The lift does not take the skiers up to the bar.

9. Height reached by plane = 26 × tan(9°) ≈ 4.12

The plane flies under the cloud therefore;

Eric will be able to see his Dad's plane.

10. Length required for the metal brace = 21/(cos(73°)) ≈ 71.83

Therefore;

The brace that wastes less metal is the 73 m. brace, because the 71 m. brace is not long enough and two braces will be required.

Learn more about trigonometric ratios here:

brainly.com/question/24349828

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7 0

2 years ago

Erin and steve divided 0.84

umka2103 [35]

Answer:

Step-by-step explanation:

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5 0

4 years ago