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3 years ago

A/-8+15greater than 23

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

Answer: $a$

Step-by-step explanation:

$\frac{a}{-8}+15>23$

Subtract 15 from both sides.

$\frac{a}{-8}+15-15>23-15\\\frac{a}{-8} >8$

Multiply by -8 to isolate a.

NOTE: When you multiply or divide by negative numbers, the inequality sign flips.

$-8*\frac{a}{-8}>8*-8$

$a$

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Solve the following non-linear system. Show all work.

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Answer:

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Step-by-step explanation:

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3 years ago

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A=1/2b(c+d) solve for c

TiliK225 [7]

Answer:

c = $\frac{2A}{b}$ - d

Step-by-step explanation:

Multiply both sides by 2 to eliminate the fraction

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3 years ago

3. Bob has 8 pieces of square index cards that each measures 8 inches per side. Using a pair of scissors, he cuts each piece in

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The number of pieces of card stock which would result upon the process repetition is; 256 pieces.

<h3>What is the number of card stock pieces resulting from the cutting process?</h3>

According to the task content, it follows that since, there are 8 pieces of the 8 inches per side cards initially, the resulting number of cards after carrying out the process of cutting cards into half five, 5 times is;

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1 year ago

Need help please!! Will give points

Nataly [62]

Answer:

1. $\bar X=5.10$

2. $\sigma=0.47$

3. $\sigma^2=0.22$

Step-by-step explanation:

<u>QUESTION 1</u>

The given data set for the expenditure is

$4.85,5.10,5.50,4.75,4.50,5.00,6.00$

The formula for calculating the mean is given by,

$\bar X=\frac{\sum x}{n}$

We need to add all the expenditure and divide by the total number of days.

$\bar X=\frac{4.85+5.10+5.50+4.75+4.50+5.00+6.00}{7}$

This gives us,

$\bar X=\frac{35.7}{7}$

$\Rightarrow \bar X=5.10$ to the nearest hundredth.

<u>QUESTION 2</u>

The standard deviation of the data set is given by the formula;

$\sigma =\sqrt{\frac{\sum(x-\bar X)^2}{n} }$

This implies that,

$\sigma =\sqrt{\frac{(4.85-5.10)^2+(5.10-5.10)^2+(5.50-5.10)^2+(4.75-5.10)^2+(4.50-5.10)^2+(5.00-5.10)^2+(6.00-5.10)^2}{7} }$

This will give us,

$\sigma =\sqrt{\frac{(-0.25)^2+(0.00)^2+(0.40)^2+(-0.35)^2+(-0.60)^2+(-0.10)^2+(0.90)^2}{7} }$

$\sigma =\sqrt{\frac{0.0625+0+0.16+0.1225+0.36+0.01+0.81}{7} }$

$\sigma =\sqrt{\frac{61}{280} }$

$\sigma =0.466775$

to the nearest hundredth,

$\sigma =0.47$ .

<u>QUESTION 3</u>

The variance is the square of the standard deviation.

$\sigma^2 =(0.46675)^2$

$\sigma^2 =0.217857$

To the nearest hundred gives,

$\sigma^2 =0.22$

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3 years ago

In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω

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Answer:

Acceleration of the particle = $bw^{2}$

Step-by-step explanation:

We are given the position vector of a particle moving in a circle of radius b units.

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Velocity , v = $\frac{dr}{dt}$ = -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v = $\sqrt{v_x^{2} +v_y^{2} }$

Squaring both sides,

$v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))$

Since $sin^{2}(wt)+cos^{2}(wt))$ = 1

$v^{2} = b^{2}w^{2}$

The acceleration towards the centre is called the centripetal acceleration and is given by

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2 years ago