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3 years ago

What is the following sum? 4(^5sqrtx^2y) + 3(^5sqrtx^2y)

Mathematics

2 answers:

Schach [20]3 years ago

7 0

Answer:

option 3

Step-by-step explanation:

$(4 + 3) \sqrt[5]{x {}^{2}y } = 7 \sqrt[5]{ {x}^{2}y }$

Sophie [7]3 years ago

4 0

Answer:

Step-by-step explanation:

$4( \sqrt[5]{ {x}^{2}y } ) + 3( \sqrt[5]{ {x}^{2} y} )$

$(taking \: \sqrt[5]{ {x}^{2}y } \: as \: common \: then)$

$= \sqrt[5]{ {x}^{2} y} (4 + 3)$

$= \sqrt[5]{ {x}^{2} y} \times 7$

$= 7 (\sqrt[5]{ {x}^{2}y } )(ans)$

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What is the slope intercept equation for this graph someone answer the question pls

disa [49]

Answer:

-2,3

Step-by-step explanation:

7 0

3 years ago

solve for b 1/4(b-8)=2. I hope you have time to answer this. Thank you so much for your time you guys make me so happy

Mademuasel [1]

Answer:

Step-by-step explanation:

1/4(b-8)=2

b-8=2/(1/4)

b-8=2(4/1)

b-8=8

b=8+8

b=16

8 0

3 years ago

Read 2 more answers

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

For her exercise today, Keiko plans to both run and swim. Let r be the number of laps she runs and let s be the number of laps s

DerKrebs [107]

Answer:

5r + 3s ≥ 30

Step-by-step explanation:

We know Keiko wants to do exercise at least 30 minutes a day, and she wants ideally to mix up the activities, by combining running and swimming into her schedule.

Since each lap she runs takes 5 minutes and each lap she swims takes 3 minutes, these will be the left side of the inequality:

5r + 3s

She wants to exercise at least 30 minutes a day, so...

5r + 3s ≥ 30

she could do 3 laps of running and 5 laps of swimming for example.

6 0

3 years ago

Which of these statements best describes the relation shown in Item 1?

Misha Larkins [42]

Answer:

It is a many-to-one relation

Step-by-step explanation:

Given

See attachment for relation

Required

What type of function is it?

The relation can be represented as:

$\left[\begin{array}{c}x\\ \\-5\\-2\\1\\5\end{array}\right]$ $\left[\begin{array}{c}y\\ \\10\\11\\4\\10\end{array}\right]$

Where

$x = domain$ and $y = range$

Notice that the range has an occurrence of 10 (twice)

i.e.

$(x_1,y_1) = (-5,10)$ and $(x_2,y_2) = (5,10)$

In function and relations, when two different values in the domain point to the same value in the range implies that, <em>the relation is many to one.</em>

8 0

3 years ago