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3 years ago

Help please, it’s due tomorrow!! :/

Mathematics

2 answers:

Dovator [93]3 years ago

8 0

I’m in highschool and ion even know this.

Murljashka [212]3 years ago

6 0

Answer: to graph them just find it on the number line and the greatest absolute value is -10 and the absolute value is 10 and the two numbers with the same absolute value is 6 and -6 with the absolute value of 6

-7<-1 |-7|>|-1| 0>-1 |0|<|-1| 12=|12| -12=-|12|

Step-by-step explanation:

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Ilia_Sergeevich [38]

Albert has ridden 1.3 mi

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4 years ago

Find all solutions in the interval [0, 2π).<br> sec2 x - 2 = tan2 x

maks197457 [2]

The equation is equivalent to 1/cos(2x)-2=sin(2x)/cos(2x). Make the denominator cos(2x) for all terms, we get 1-2cos(2x)=sin(2x). Since cos^2(2x)+sin^2(2x)=1, let cos(2x)=a, substitute sin(2x) by +-\sqrt(1-a^2), and solve the equation with only one variable a. We get a=4/5 or 0, but cos(2x) cannot be 0, otherwise sec(2x) is undefined. 2x=36.9 degrees, so x=18.9 degrees approximately.

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3 years ago

What is the value of c?

Ksenya-84 [330]

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7 0

3 years ago

Which expression is equal to 2/sqrt7

Amanda [17]

The one on the top is correct

Step-by-step explanation:

4 0

3 years ago

There are three different hoses used to fill a pool: hose x, hose y, and hose z. Hose x can fill the pool in a days, hose y in b

WARRIOR [948]

Answer:

C) I and III only

Step-by-step explanation:

Let full pool is denoted by O

Days Hose x takes to fill pool O = a

Pool filled in one day x = O/a

Days Hose y takes to fill pool O = b

Pool filled in one day y = O/b

Days Hose z takes to fill pool O = c

Pool filled in one day z = O/c

It is given that

a>b>c

$a>b>c>d\\\implies x$

Days if if x+y+z fill the pool together = d

1 day if x+y+z fill the pool together $=O(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{O}{d}---(1)$

I) d < c

d are days when hose x, y, z are used together where as c are days when only z is used so number of days when three hoses are used together must be less than c when only z hose is used. So d < c

III) $\frac{c}{3}$

Using (1)

$\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad(a>b>c)\\(ab+bc+ca)\frac{abc}{3ab}\\\\d>\frac{c}{3}$

Similarly

$\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad a>b>c\\(ab+bc+ca)>3bc\\\\d=\frac{abc}{ab+bc+ca}$

So,

$\frac{c}{3}$

3 0

3 years ago