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3 years ago

Which of the following are true statements about the expression 12^5

Mathematics

1 answer:

Usimov [2.4K]3 years ago

4 0

12×12×12×12×12 is equivalent to 12^5 (a^b means a is to be written down B times all being multiplied.)

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The double number line shows that 3 back-to-school packages contain 36 pencils.

lord [1]

Answer:

The answer is A

Step-by-step explanation:

i just got it right

7 0

3 years ago

A magazine subscription costs $29.88 for 12 issues or $15.24 for 6 issues. Which subscription costs more per issue? How much mor

Valentin [98]

Answer:

The magazine that cost $15.24 for 6 issues cost more by $0.60/60 cents.

Step-by-step explanation:

Since $15.24 is 6 issues each and you need to even out the issues add $15.24+$15.24 then you would get $30.48, because its 15.24 each 6 add them both and it would be 12 issues. And the first issue $29.88 for 12 issues is only $29.88 while the other is $30.48 so the one with 6 issues/the second cost more than the first.

60 cents more because 29.88+60=30.48.

3 0

3 years ago

each blade on a windmill is in the shape of a triangle with a base of 7 feet and a height of 4 ft if there are four blades on th

Sunny_sXe [5.5K]

The total area of the blades is the area of a triangle multiplied by 4.

We have then:

$A = (4) * (1/2) * (b) * (h)$

Where,

b: base of the blades

h: height of the blades

Substituting values we have:

$A = (4) * (1/2) * (7) * (4)$

Doing the calculation we have that the area of the blades is given by:

$A = 56 feet ^ 2$

Answer:

The total area of the blades is:

$A = 56 feet ^ 2$

6 0

3 years ago

Read 2 more answers

It took Everly 55 minutes to run a 10-kilometer race last weekend. If you know that 1 kilometer equals 0.621 mile, how many minu

slega [8]

$\bf \cfrac{10km}{55mins}\cdot \cfrac{0.621miles}{1km}\implies \cfrac{62.1miles}{55mins}\approx 1.12909909\approx \boxed{1.13\frac{miles}{mins}}$

5 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers