9. Which unit of measure would be appropriate for the volume of a cube with sides of 2 meters.​

Answer:

OOOO Shoot im not in high school

Step-by-step explanation:

Answer:

8n² - 5

Step-by-step explanation:

n = 'a number'

The expression is  8x² - 5.

A mathematical expression is a phrase that has a minimum of two numbers or variables and at least one mathematical operation. Let's examine the writing of expressions. The other number is x, and a number is 6 greater than half of it. In a mathematical expression, this proposition is denoted by the expression x/2 + 6.

Given:

The product of eight and a number squared decreased by 5

let the number be x.

product of eight and a number squared

= 8x²

and decreased by 5

= 8x² - 5

hence, the expression is  8x² - 5.

Learn more about expression here:

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#SPJ2

Answer:

25.)Yesss

26.)Yesss she have enough money to buy a popcorn and soda

27.)?

Step-by-step explanation:

#Hopeee it Helpssss

follow me pls.

Answer:

Hey!

Step-by-step explanation:

nnnnnnnbbbbbbbbbbbvvvgbhjnn,,

Answer:

Step-by-step explanation:

Answer: -15

Step-by-step explanation: in the picture

Step-by-step explanation:

Basically,

expand the brackets; multiply everything in the bracket by -3,

-3(x+5)=-3x+-15

= -3x-15

There ya go, hope this helps...

Answer:

7

Step-by-step explanation:

Answer:

Answer: 8PM and 8AM

Step-by-step explanation:

2 |x-2| -12 = 0

2 |x-2| = 12

|x-2| = 6

x-2 = 6

x= 8

12 noon + 8 Hrs

= 8PM

x-2 = -6

x = -4

12 noon - 4 hrs

= 8AM

Given: [tex]cosec[/tex] [tex]A[/tex]  [tex]=\sqrt{2}[/tex]

∴ [tex]A=\frac{\pi }{4} =45^{0}[/tex]

[tex]sin[/tex] [tex]A[/tex] [tex]=sin[/tex] [tex]45^{0}[/tex] = [tex]\frac{1}\sqrt{2}[/tex]

[tex]cot[/tex] [tex]A[/tex] [tex]=cot[/tex] [tex]45^{0}[/tex] [tex]=1,[/tex]

[tex]tan[/tex] [tex]A[/tex] [tex]=[/tex] [tex]tan[/tex] [tex]45^{0}[/tex] [tex]=1[/tex]

[tex]cos[/tex] [tex]A[/tex] [tex]=[/tex] [tex]cot[/tex] [tex]45^{0}[/tex] [tex]=\frac{1}{\sqrt{2} }[/tex]

∴ [tex]=2[/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

METHOD 1: Using Trigonometric values.

We have given that,

[tex]\sf\cosec\theta=\sqrt2[/tex]

So,

[tex]\sf\theta=45^{\circ}[/tex]

We need to find:

[tex]\sf\dfrac{1+\sin^2\theta+\cos^2\theta}{1+\sec^2\theta-\tan^2\theta}[/tex]

So,

[tex]\sf\longmapsto\dfrac{1+\sin^2(45^{\circ})+\cos^2(45^{\circ})}{1+\sec^2(45^{\circ})-\tan^2(45^{\circ})}[/tex]

[tex]\sf\longmapsto\dfrac{1+\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2}{1+(\sqrt2)^2-(1)^2}[/tex]

[tex]\sf\longmapsto\dfrac{1+\frac{1}{2}+\frac{1}{2}}{1+2-1}[/tex]

[tex]\sf\longmapsto\dfrac{1+1}{2}[/tex]

[tex]\sf\longmapsto\dfrac{2}{2}[/tex]

[tex]\sf\longmapsto\bf1[/tex]

METHOD 2: Using Trigonometric Identities

We need to find:

[tex]\implies\dfrac{1+\sin^2\theta+\cos^2\theta}{1+\sec^2\theta-\tan^2\theta}[/tex]

But, we know that,

[tex]\sf\red⇛ \sin^2\phi+\cos^2\phi=1[/tex]

And,

[tex]\sf\red⇛ \sec^2\phi-\tan^2\phi=1[/tex]

So,

[tex]\sf\longmapsto\dfrac{1+(\sin^2\theta+\cos^2\theta)}{1+(\sec^2\theta-\tan^2\theta)}[/tex]

[tex]\sf\longmapsto\dfrac{1+1}{1+1}[/tex]

[tex]\sf\longmapsto\dfrac{2}{2}[/tex]

[tex]\sf\longmapsto\bf1[/tex]

Therefore, by both methods the answer is 1.

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Similar questions

Prove that:

\cos( \frac{3\pi}{2} + \theta) \cos(2\pi + \theta) [ \cot( \frac{3\pi}{2} - \theta) + \cot(2\pi + \theta) ] = 1...

https://brainly.com/question/25207676

Step-by-step explanation:

2+2 = 4 - 6= -2+8=6+2=8

Answer: 8

Answer:

48

Step-by-step explanation:

First add 2 + 2 = 4

Next subtract 4 - negative 6 = -2

Then add -2 + 8 = 6

Last multiply 6 * 8 = 48

Answer:

2n

Step-by-step explanation:

the output will be 2n

please brainliest

1. C because 3×3=9

9×3=27

27×3=81.....

2. Answer is C explaination in picture number 2

3. D explaination in picture number 1

Answer:

3/4 I think its that

Step-by-step explanation:

its a guess

Answer:

its the lastone

Step-by-step explanation:

Answer:

y ≤ 20-8x

Step-by-step explanation:

Darren needs to spend no more than 15 hours on cardio exercise (y). To figure out how much time he has we need to do 20-8x. Any leftover hours are the hours left for him to spend doing his exercises. Since y has to be less than or equal to 20-8x, the system of linear inequalities should be:

y ≤ 20-8x

Answer:

It is 2 π r

Step-by-step explanation:

Hope this helps you!!!1

            4^t × 2^t = 2^what         [convert all terms to the same base (2)]

∴   2^t × 2^t × 2^t = 2^what       [use the multiplicat law of exponents]

              2^(t+t+t) = 2^what       [simplify]

                   2^3t =  2^what

Answer:

95.06506849315

Step-by-step explanation:

Basically 95 years old which is very old Pls brainliest Thx! :D

Answer:

lets see according to my cackulations.... suzie would be, 6.....she would be six feet under the ground. dead. deceaced.gone.

Step-by-step explanation:

Answer:

9/2 or 4 1/2

Step-by-step explanation:

Therefore, the answer is 9/2 or 4 1/2.

Answer: 3

Step-by-step explanation:

3/4 ÷ 1/6

3/4 × 6/1

Divide 4 and 6 by 2

3/2 × 3/1

3 × 3 / 2 × 1

6 / 2 = 3

Answer:

he prepared for 18 plants altogether

Answer:

The circumference is the distance around a circle. In other words, it's the perimeter of the circle. And we find the circumference by using the formula C = 2πr.

Step-by-step explanation:

Answer: 10h+ 4

Step-by-

It will take twelve (12) plumbers 9 weeks to install the same number of lines

From the question,

We have that, 18 plumbers can install 540 water lines in 6 weeks,

To determine how long it will take 12 plumbers to install the same number of lines,

First, we will determine how long it will take 1 plumber.

If 18 plumbers can install 540 water lines in 6 weeks

Then,

1 plumber will install 540 water lines in 18 × 6 weeks

18 × 6 weeks = 108 weeks

∴ It will take 1 plumber 108 weeks to install 540 water lines

Now, for the number of weeks it will take 12 plumbers

If 1 plumber can install 540 water lines in 108 weeks

Then,

12 plumbers will install 540 water lines in [tex]\frac{108}{12}[/tex] weeks

[tex]\frac{108}{12} = 9 \ weeks[/tex]

Hence, it will take twelve (12) plumbers 9 weeks to install the same number of lines

Learn more here: https://brainly.com/question/18880656

The system of inequalities models correctly the number of cookies is c + b ≥ 125 and 8c + 12b ≤ 2000

Let c represent the number of cookies and let b represent the number of brownies.

Since Genie want to make more than 125 cookies and brownies altogether, hence this is given by:

c + b ≥ 125   (1)

Also, Genie has a total of 2,000 grams of flour, hence:

8c + 12b ≤ 2000  (2)

Therefore the system of inequalities models correctly the number of cookies is c + b ≥ 125 and 8c + 12b ≤ 2000

Find out more at: https://brainly.com/question/15137133?referrer=searchResults

Answer:

35

Step-by-step explanation:

70 minutes for 24 songs

70/2 minutes for 24/2 songs

35 minutes for 12 songs.

Answer:

35 minutes for 12 songs

Step-by-step explanation:

hope this helps

Answer:

1) [tex]y=\frac{3}{2}x+8[/tex]

2)[tex]y=\frac{13}{11}x +\frac{12}{11}[/tex]

3)[tex]y=\frac{9}{7}x+1[/tex]

4)[tex]y=\frac{1}{3}x-2[/tex]

5)[tex]y=-\frac{6}{5}x-3[/tex]

6)[tex]y=4x-1[/tex]

7)[tex]y=\frac{11}{4}x-8[/tex]

8)[tex]y=\frac{11}{8}x+6[/tex]

Answer:

[tex]\sf 1) 3x-2y=-16[/tex]

subtract both sides by 3x:

[tex]\sf -2y=-16-3x[/tex]

[tex]\sf \cfrac{-2y}{-2} =\cfrac{3x-16}{-2}[/tex]

[tex]\boxed {\sf y=\frac{3}{2}x +8}[/tex]

________________

[tex]\sf 2) 13x-11y=-12[/tex]

Subtract 13x from both sides:

[tex]\sf 13x+11y-13x=-12-13x[/tex]

[tex]\sf \cfrac{-11y}{-11} =\cfrac{-12-13x}{-11-11}[/tex]

[tex]\boxed {\sf y=\cfrac{13}{11} \:x+\cfrac{12}{11} }[/tex]

___________________

[tex]\sf 3) 9x-7y=-7[/tex]

Subtract 9x from both sides:

[tex]\sf \cfrac{-7y}{-7} =\cfrac{9x-7}{-7}[/tex]

[tex]\boxed {\sf y=\cfrac{9}{7} \:x+1}[/tex]

___________________

[tex]\sf 4)x-3y=6[/tex]

Add 3y from both sides:

[tex]\sf x-3y+3y=6+3y[/tex]

[tex]\sf x=6+3y[/tex]

Subtract 6 from both sides:

[tex]\sf x-6=3y[/tex]

Divide both sides by 3:

[tex]\sf \cfrac{x}{3}-\cfrac{6}{3} =\cfrac{3y}{3}[/tex]

[tex]\sf \cfrac{1}{3} \:x-2=y[/tex]

[tex]\boxed {\sf y=\cfrac{1}{3} \:x-2}[/tex]

____________________

[tex]\sf 5) 6x+5y=-15[/tex]

Subtract 6x from both sides:

[tex]\sf \cfrac{5y}{5}=\cfrac{-6x-15}{5}[/tex]

[tex]\boxed {\sf y=-\cfrac{6}{5} \:x-3 }[/tex]

_____________________

[tex]\sf 6) 4x-y=1[/tex]

Subtract 4x from both sides:

[tex]\sf \cfrac{-y}{-1} =\cfrac{-4x+1}{-1}[/tex]

[tex]\boxed {\sf y=4x-1}[/tex]

_______________________

[tex]\sf 7)11x-4y=32[/tex]

Subtract 11x from both sides:

[tex]\sf \cfrac{-4y}{4} =\cfrac{-11x+32}{-4}[/tex]

[tex]\boxed {\sf y=\frac{11}{4} \:x-8}[/tex]

_______________________

[tex]\sf 8) 11x-8y=-48[/tex]

Subtract 11x from both sides:

[tex]\sf \cfrac{-8y}{-8} =\cfrac{11-48}{-8}[/tex]

[tex]\boxed {\sf y=\frac{11}{8} \:x+6}[/tex]

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