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4 years ago

Can someone tell me the answer?

Mathematics

1 answer:

Dmitry_Shevchenko [17]4 years ago

3 0

Answer:

the first one has one solution because eventually they will cross

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Select the correct answer. Simplify the expression to a single numerical value.

KIM [24]

432

I know this bc if you do the equation is gonna be 3•8•2•9 then it will be 24•2•9 then it’s gonna be 48•9 to get 432 as your answer

7 0

3 years ago

Read 2 more answers

The figures are similar. The area of one figure is given. Find the area of the other figure to the nearest whole number.

Akimi4 [234]

Answer: Option b.

Step-by-step explanation:

First, you need to calculate the ratio of the area. This is:

$ratio=(\frac{64}{28})^2\\\\ratio=\frac{256}{49}$

You know that the area of the smaller trapezoid is 771 m², then you can set up the following proportion, where "x" is the area of the larger trapezoid. Then you have:

$\frac{256}{49}=\frac{x}{771}$

Now you must solve for "x". Therefore, you get that the area of the larger trapezoid is:

$x=(771)(\frac{256}{49})\\\\x=4,028m^2$

3 0

4 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

Please help ASAP ....

vlada-n [284]

Answer:

21 3/4

Step-by-step explanation:

3 times 7 1/4 since the scale is 1:7 1/4. Good luck!! :)

7 0

3 years ago

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The Coolmathz family is ordering pizza. They order two pizzas. One is $15, and the other is $12. They use a coupon for 5% off th

DaniilM [7]

Answer:

30.51$

Step-by-step explanation:

15+12=27$

5% of 27$=1.35$

18% of 27=4.86

Do the math: 27+4.86-1.35=30.51

6 0

3 years ago