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3 years ago

FINAL. F.RE.E .P.O.IN.T.S!!!!!!!

Mathematics

2 answers:

Semmy [17]3 years ago

5 0

Jdsjsbhssisihqbbjsoxoodjsbdvd

Strike441 [17]3 years ago

4 0

Answer:

Thank you very much for the points!!

You might be interested in

Show all work and receive brainliest!

Svet_ta [14]

Answer:

Lower Quartile: 62

Upper Quartile: 81

Interquartile Range: 19

Step-by-step explanation:

To find the lower quartile, you want to find the median from the minimum to the median.

49, 55, 62, 64, 67

The median of this is 62. Therefore, 62 is the lower quartile.

To find the upper quartile, you want to find the median from the median to the maximum.

76, 79, 81, 82, 83

The median of this is 81. Therefore, 81 is the upper quartile.

To find the interquartile range, you subtract the upper and lower quartile.

81-62=19

The difference is 19. Therefore, the interquartile range is 19.

5 0

3 years ago

Find the area of the region y=4cos(x), y=4-4cos(x), 0

kari74 [83]

Answer:

A= 4 $\pi$ sq. unit

Step-by-step explanation:

4 0

3 years ago

Pls hurry!!!!!!!!!!!!!!!!!!!!!!!!!

Vsevolod [243]

Answer:

It's B.

Step-by-step explanation:

6 0

3 years ago

According to the rational root theron what are all the potential roots of f(x)=9x^4-2x^2-3x+4

Viefleur [7K]

Answer:

Potential roots: $\frac{9}{4},\frac{9}{2},9,\frac{3}{4},\frac{3}{2},3, \frac{1}{4},\frac{1}{2},1$

Step-by-step explanation:

Simply put, the rational roots theorem tells us that if there are any rational roots of a polynomial function, they must be in the form

± $\frac{FactorsOfa_{0}}{FactorsOfa_n}$

Where

a_n is the number before the highest power of the polynomial, and

a_0 is the constant in the polynomial

From the polynomial shown, we have a_n = 9 and a_0 = 4

The factors of 9 are 9, 3, 1

and

The factors of 4 are 4,2,1

So, if there are any rational roots, they would be:

± $\frac{FactorsOfa_{0}}{FactorsOfa_n}$

± $\frac{9,3,1}{4,2,1}$

Which is ± 9/4, 9/2, 9/1, 3/4, 3/2, 3/1, 1/4, 1/2, 1/1

$\frac{9}{4},\frac{9}{2},9,\frac{3}{4},\frac{3}{2},3, \frac{1}{4},\frac{1}{2},1$

5 0

3 years ago

The intensity of a light source at a distance is directly proportional to the strength of the source and inversely proportional

jolli1 [7]

Answer:

x= $\frac{16}{\sqrt[3]{2}+1 }$

Step-by-step explanation:

Q= illumination

I = intensity

Q= I/d^2

Q_total = $\frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}$

= $\frac{I}{x^2}+\frac{2I}{(16-x)^2}$

now Q' = 0

⇒I ${-\frac{2}{x^3}}+\frac{4}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

[/tex] $\frac{1}{x^3} = \frac{2}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

is the required point

5 0

3 years ago