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3 years ago

A portion of the Quadratic Formula proof is shown. Fill in the missing statement.

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

4 0

Answer:

Option A.

Step-by-step explanation:

we have

$x+\frac{b}{2a} =\pm\sqrt{\frac{b^2-4ac}{4a^2}}$

Simplify the right side of the equation

we know that

$\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\pm\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}=\pm\frac{\sqrt{b^2-4ac}}{2a}$

The expression is

$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$

therefore

x plus b over 2 times a equals plus or minus the square root of b squared minus 4 times a times c, all over 2 times a

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In front of a store, there is a row of parking spaces. Cars park parallel to one another, with the front of each car facing the

Gelneren [198K]

The width used for the car spaces are taken as a multiples of the width of

the compact car spaces.

Correct response:

The store owners are incorrect

<h3 /><h3>Methods used to obtain the above response</h3>

Let <em>x</em><em> </em>represent the width of the cars parked compact, and let a·x represent the width of cars parked in full size spaces.

We have;

Initial space occupied = 10·x + 12·(a·x) = x·(10 + 12·a)

New space design = 16·x + 9×(a·x) = x·(16 + 9·a)

When the dimensions of the initial and new arrangement are equal, we have;

10 + 12·a = 16 + 9·a

12·a - 9·a = 16 - 10 = 6

3·a = 6

a = 6 ÷ 3 = 2

a = 2

Whereby the factor <em>a</em> < 2, such that the width of the full size space is less than twice the width of the compact spaces, by testing, we have;

10 + 12·a < 16 + 9·a

Which gives;

x·(10 + 12·a) < x·(16 + 9·a)

Therefore;

The initial total car park space is less than the space required for 16

compact spaces and 9 full size spaces, therefore; the store owners are

incorrect.

Learn more about writing expressions here:

brainly.com/question/551090

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2 years ago

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

3 years ago

If 30% of a number, n, is 120, what is 82% of n?

Naya [18.7K]

Answer:

1.28

Step-by-step explanation:

I did the math.

5 0

2 years ago

Read 2 more answers

PLEASE HELP TRYING TO MAKE HONOR ROLL

maria [59]

Answer:

Hulian's age is 7.

Thomas's age is 22.

Step-by-step explanation:

Let Hulian = h

Let Thomas = t

Set the system of equation:

h = t - 15

h + t = 29

Plug in t - 15 for h in the second equation:

(t - 15) + t = 29

Simplify. Combine like terms:

2t - 15 = 29

Isolate the variable, t. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS. First, add 15 to both sides:

2t - 15 (+15) = 29 (+15)

2t = 44

Divide 2 from both sides:

(2t)/2 = (44)2

t = 44/2

t = 22

Plug in 22 for t in one of the equations:

h = t - 15

h = 22 - 15

h = 7

Hulian's age is 7.

Thomas's age is 22.

7 0

3 years ago

I need help on number 8 and please show how you solved it :))

dmitriy555 [2]

Answer:

32.04% to 2 decimal places.

Step-by-step explanation:

As a fraction it is 17.3 / 54

To get the percentage we multiply by 100:

= (17.3/54) * 100

= 32.04%

8 0

3 years ago