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3 years ago

If x>0, what values of c and d make the equations true?

Mathematics

2 answers:

Dominik [7]3 years ago

6 0

Answer:

a=7

b=5

Step-by-step explanation:

pantera1 [17]3 years ago

5 0

Answer:

In equation A, c = 7

In equation B, d = 5

Step-by-step explanation:

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The diagonals of a kite have lengths of x inches and 18x inches. The area of the kite is given by A=9x^2. Identify the equation(

Firdavs [7]

Answer:

Step-by-step explanation:

5 0

3 years ago

HELP PLS NEED THIS IN 5 MIN

swat32

Answer:

0.25

Step-by-step explanation:

1. Add up all of the students.

20+20+25+7+13+5+3+2+5=100

2. Make a ratio of that number and the number of all the students in Music/Drama

MUSIC/DRAMA: 7+13+5=25

$\frac{25}{100}$

3. Simplify

$\frac{25}{100} = \frac{5}{20} =\frac{1}{4}$ = 0.25

8 0

3 years ago

If the diameter of his flower bed is 100 inches what is distance around the garden

Gala2k [10]

Well if pi is referred to as 3.14 then the distance around the garden would be 314.
The formula for Circumfrence is pi times diameter, so 3.14 times 100 = 314 (circumfrence)

8 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

2 years ago

Read 2 more answers

Mr. Wilson invested money in two accounts. His total investment was $20,000. If one account pays 5% in interest and the other pa

blagie [28]

a = amount invested at 5%

b = amount invested at 2%

now, we know the total invested by Mr Wilson was 20000, so whatever "a" and "b" might be, we know that a + b = 20000.

$a + b = 20000\implies b = 20000-a \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{5\% of a}}{\left( \cfrac{5}{100} \right)a}\implies 0.05a~\hfill \stackrel{\textit{2\% of b}}{\left( \cfrac{2}{100} \right)a}\implies \begin{array}{llll} 0.02b\\\\ \stackrel{substituting}{0.02(20000-a)} \end{array}$

now, we know the total of earned interest is 550 bucks, so then

$0.05a~~ + ~~0.02(20000-a)~~ = ~~550\implies 0.05a+400-0.02a=550 \\\\\\ 0.03a+400=550\implies 0.03a=150\implies a=\cfrac{150}{0.03}\implies a=5000 \\\\\\ ~\hspace{24em}\stackrel{20000~~ - ~~5000}{b=15000}$

5 0

1 year ago