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ad-work [718]
3 years ago
11

Can someone help with this please?

Mathematics
2 answers:
ipn [44]3 years ago
3 0

7^2=7\cdot7=49\\\\8^2=8\cdot8=64\\\\\sqrt{49}

alex41 [277]3 years ago
3 0

Answer:

Its 52

Step-by-step explanation:

the square root of 52 is 7.2

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What is absolute extrema of cube root of x on I=[-3,8]
hichkok12 [17]
<span>These are points where f ' = 0. Use the quiotent rule to find f '. 

f ' (x) = [(x^3+2)(1) - (x)(3x^2)] / (x^3+2)^2 
f ' (x) = (2 - 2x^3) / (x^3 + 2)^2 

Set f ' (x) = 0 and solve for x. 

f ' (x) = 0 = (2-2x^3) / (x^3+2)^2 

Multiply both sides by (x^3+2)^2 

(x^3+2)^2 * 0 = (x^3+2)^2 * [(2-2x^3)/(x^3+2)^2] 
0 = 2 - 2x^3 

Add 2x^3 to both sides 

2x^3 + 0 = 2x^3 + 2 - 2x^3 
2x^3 = 2 

Divide both sides by 2 

2x^3 / 2 = 2 / 2 
x^3 = 1 

Take cube roots of both sides 

cube root (x^3) = cube root (1) 
x = 1. This is our critical point 

2) Points where f ' does not exist. 

We know f ' (x) = (2-2x^3) / (x^3+2)^2 

You cannot divide by 0 ever so f ' does not exist where the denominator equals 0 

(x^3 + 2)^2 = 0. Take square roots of both sides 
sqrt((x^3+2)^2) = sqrt(0) 
x^3 + 2 = 0. Add -2 to both sides. 
-2 + x^3 + 2 = -2 + 0 
x^3 = -2. Take cube roots of both sides. 
cube root (x^3) = cube root (-2) 
x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point. 

3) End points of the domain. 

The domain was clearly stated as [0, 2]. The end points are 0 and 2. 

Therefore, our only options are: 0, 1, 2. 

Check the intervals 

[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign. 

In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive. 

In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative. 

Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum. 

f (0) = 0 
f (1) = 1/3 
f (2) = 1/5 

Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute
maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
3 0
4 years ago
2/4=a/2 what is it algebra 1
harkovskaia [24]

if you're asking for the solution of this equation in algebra, the answer will be

2/4=a/2

(2/4)2=a

4/4=a

a=1

7 0
3 years ago
The weather forecast says that the probability of being cloudy tomorrow is 30% and the probability of raining is 25%. If it is c
Bumek [7]
<h3>Answer:</h3>

\large\boxed{C).\,13.5\%}

<h3>Step-by-step explanation:</h3>

In this question, we're trying to find the probability of it being cloudy and raining.

In this case, we know that:

  • Probability of it being cloudy is 30%
  • Probability of it raining is 25% (this is necessarily not needed)
  • If it's cloud, the probability of it raining is 45%

With the information above, we can find the probability.

We know that from a 100% scale, the chance of it being cloudy is 30%.

We know that if it's cloudy, the chances of raining is 45%

To find the probability of it being cloudy and raining, we would multiply 0.3 (30%) by 0.45 (45%)

Solve:

0.3*0.45=0.135\\\\\text{Move the decimal place two times to the right}\\\text{to turn it into a percent}\\\\\boxed{13.5\%}

Your answer would be C). 13.5%

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
6 0
3 years ago
A package of 5 pairs of socks costs $35.95. What is the unit price of the pairs of socks?
Juliette [100K]

Answer:

7.19 a pair

Step-by-step explanation:

To find unit price divide the total by the amount

for ex

5 pairs cost 35.95 so 35.95 is the price of 5 pairs. To find out the price of 1 pair divide the price by the items or the 5 pairs of socks. 35.95 / 5 is 7.19

im sorry for my lack of knowing how to explain :(

7 0
3 years ago
The definition of power
kicyunya [14]

Answer:

When you look in the dictinary it says this...*pick any they all will get the answer correct!

Step-by-step explanation:

definition for power in math!

In mathematics, power refers to the number arrived at by raising a number to an exponent.

brainliest? hope this helps!

8 0
3 years ago
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