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2 years ago

|x+1|>8 absolute value equations and inequalities

Mathematics

1 answer:

Pepsi [2]2 years ago

7 0

x < -9 or x > 7

x + 1 < -8

x < -9

x + 1 > 8

x > 7

... when the inequality sign is facing the x, x is GREATER and it resembles an OR statement. I personally like to think of it as: greater than (greatOR than) = or statement; less than (less thAND) = and statement.

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(7y is supposed to be 7x)

Andrei [34K]

Answer:

Step-by-step explanation:

Angle E is (7x+27) because it is corresponding.

So, 7x+27+15x-5=180 (angle on a straight line is 180 degrees

22x+22=180

22x=158

x= 158/22 = 7.18'

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1 year ago

You can buy 4 apples at stop and shop for 0.96. you can buy 6 of the same apples at path mark for $1.50 which store has the bett

monitta

Stop and Shop has the better deal, as it is 0.24 cents for each apple if you divide 0.96 by 4, while Path Mark is 0.25 cents.

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2 years ago

Need help on this baldy plzz anyone.♥

AURORKA [14]

The answer is 24
subtract 156 from 180
108-156=24

5 0

2 years ago

Read 2 more answers

Can someone help me pls

Leviafan [203]

Answer:

Step-by-step explanation:

0.75 feet / day [ 1 day / 24 hours] * [12 inches / 1 foot]

This process is called dimensional analysis. The trick it uses is to cancel out units (like feet or days ) where possible. In this case, 1 day is going to cancel out and leave behind 24 hours. 1 foot will cancel out and leave behind 12 inches.

It is well worth learning how to do this, because the process is done in the higher sciences.

So what you are left with is

0.75 * 12 inches / 24 hours

0.75 inches /hour * 1/2

0.375 inches / hour.

6 0

2 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

2 years ago