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2 years ago

Which is the result when the equation y = -2/5x + 5/2 is converted to standard form?

Mathematics

2 answers:

finlep [7]2 years ago

5 0

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

to do away with the denominators, we can just use the LCD of all denominators and multiply both sides by it hmm, in this case is 10, the LCD of 5 and 2, so we'll use that.

$y=-\cfrac{2}{5}x+\cfrac{5}{2}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{10}}{10(y)~~ = ~~10\left( -\cfrac{2}{5}x+\cfrac{5}{2} \right)} \\\\\\ 10y=-4x+25 \implies 4x+10y=25$

mixer [17]2 years ago

4 0

Answer:

$4x+10y-25=0$

Step-by-step explanation:

$y=-\frac{2}{5} x+\frac{5}{2}$

Multiply by 10 on both sides,

$10y=-4x+25\\$

so $4x+10y-25=0$

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Which of these is an exponential parent function?

attashe74 [19]

The exponential parent function is D

3 0

3 years ago

Yolanda spends 8/23 hours per month playing soccer. Approximately how many hours does she play soccer in a year

Novosadov [1.4K]

Given:
8 2/3 hours per month
12 months in a year.

Number of hours she play soccer in a year.

8 2/3 must be converted to improper fraction.

((8*3)+2)/3 = 26/3

26/3 * 12 = (26 * 12)/3 = 312/3 = 104

Yolanda spends 104 hours in a year playing soccer.

5 0

3 years ago

Read 2 more answers

each truck in the line weigh two tons more than the truck before it. the trucks weigh a total of 32 tons.how many pounds does a

mixer [17]

If there are 4 trucks then the first truck weighs x
then x+2tons
then x+4tons
then x+6tons = 32tons
so 4x+12=32
-12 -12
/4 /4
x=5
so plug it back in
truck 1= 5tons or 10,000lbs
truck 2= 7tons or 14,000lbs
truck 3= 9tons or 18,000lbs
<span>truck 4= 11tons or 22,000lbs</span>

6 0

4 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago

Help me please ❤️❤️❤️❤️

melomori [17]

In a quadratic function, $f(x)= ax^2+bx+c$ the leading coefficient is $a$ , i.e. the coefficient of $x^2$ , and the constant term is $c$ , i.e. the coefficient without any power of $x$ .

So, if you want leading coefficient 2 and constant term -3, your function needs to be like

$f(x)=2x^2+bx-3$

for any value of $b$ . Among your options:

The first is not a quadratic (it starts with $x^3$ )
The second has leading term -3 and constant term 2
The third is not a quadratic (same as the first)
The last is correct.

4 0

3 years ago