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2 years ago

Im not sure what method is best used to solve these types of inequalities.

Mathematics

2 answers:

Harrizon [31]2 years ago

8 0

Answer:

I would say solve for the x value

Step-by-step explanation:

miskamm [114]2 years ago

6 0

Answer:

There is no specific method you can solve it as you like

Step-by-step explanation:

you can do it as a normal equation first get rid of the (x-9) by multiplying the whole radical by x-9/x-9, then you will get rid of the x-9 down and you will have (x-9) . (x+6) . (x-1) $\leq$ 0 then multiply them together and solve it normally. Also dont forget to do the check part :)

If you need further help comment down :)

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#2 please help find the correct answer.

STatiana [176]

Answer:

Mean = Mean is all of the numbers added up and divided by the amount of numbers you have added up. Therefore its 414 ÷ 15 = 27.6

Mode = Most common number = 35

Range = Largest number - Smallest number

40 - 5 = 35

Median = 32 (Middle number)

8 0

2 years ago

3×52 A. 5/6 B. 15/2 C.8/2 D.15/6

ohaa [14]

D.) 15/6

- 3 times 52 = 156, and since you can't exactly convert it to a fraction then i'm assuming that it is 'D'.

- Ouma :>

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3 years ago

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Andreyy89

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2 years ago

Water is being pumped into an inverted conical tank at some constant rate. The tank has a height of 600 cm and the diameter of t

ankoles [38]

Answer:

The water is being pumped at a speed of

$\boxed{\bf \frac{10592000*\pi}{27}\;cm^3/min}$

Step-by-step explanation:

By congruence of triangles, the radius r of the cone base when its height is 200 cm satisfies the relation

r/200 = 400/600

<h3>(See picture attached) </h3>

So, r = 400/3 cm when the water is 200 cm high.

The volume of a cone with radius of the base = R is given by

$\bf V=\frac{\pi R^2h}{3}$

So, the volume of water when it is 200 cm high is

$\bf V_1=\frac{\pi* (400/3)^2*200}{3}=\frac{32000000*\pi}{27}\;cm^3$

One minute later, the height of the water is 200 cm + 20 cm = 220 cm

The radius now satisfies

r/220 = 400/600

and now the radius of the base is

r = 440/3

and the new volume of water is

$\bf V_2=\frac{\pi* (440/3)^2*220}{3}=\frac{42592000*\pi}{27}\;cm^3$

So, the water is raising (being pumped) at a rate (speed) of

$\bf V_2-V_1=(\frac{42592000*\pi}{27}-\frac{32000000*\pi}{27})=\frac{10592000*\pi}{27}\;cm^3/min$

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2 years ago

A student constructs the angle bisectors of a triangle and then uses their intersection point as the center of a circle. The stu

Tems11 [23]

"An inscribed circle of a triangle" is the one among the following choices given in the question that the student has constructed. The correct option among all the options that are given in the question is the penultimate option or option "C". I hope that this is the answer that you were looking for and it has come to your help.

8 0

2 years ago

Read 2 more answers