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agasfer [191]
3 years ago
14

Which expression represents five times the quotient of two numbers

Mathematics
1 answer:
kotykmax [81]3 years ago
7 0

5 times the quotient of 2 numbers is 5(r/t)

Answer is the third option

5(r/t)

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PLSSSS HELP ME NOWW I DONT GET IT
kari74 [83]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
3х - 30 = у<br> 7y - 6 = 3х
mars1129 [50]

Answer:

Step-by-step explanation:

This one is simple substitution... at least, substitution is the easiest method. The first equation is 3<em>x</em> – 30 = <em>y</em>  and the second is 7<em>y</em> – 6 = 3<em>x</em>

As I look, I see 3 ways to use substitution to solve this:

  1. substitute 7<em>y</em> – 6 for 3<em>x</em>
  2. substitute 3<em>x</em> – 30 for <em>y</em>
  3. solve 3<em>x</em> – 30 = <em>y</em> for 3<em>x</em> and make it equal to 7<em>y</em> – 6

We're going to only use 1 method for the sake of time. Try the other two on your own. Assuming you don't make any mistakes, they will work.

<u>Method 1</u>:

3<em>x</em> – 30 = <em>y</em>

7<em>y</em> – 6 = 3<em>x</em>  — initial system of equations

7<em>y</em> – 6 – 30 = <em>y</em>  — substitute 7<em>y</em> – 6 for 3<em>x</em>

<u>7</u><u><em>y</em></u> – 6 – 30 = <u><em>y</em></u>  — marking like terms, bold for constants, <u>underlined</u> for variables

7<em>y</em> – 36 = <em>y</em>  — combining the constants and simplifying

Here, you could diverge into multiple paths: add 36 to both sides, subtract <em>y</em> from both sides, divide by 6 OR subtract 7<em>y</em> from both sides and divide by –6 . For the sake of time, I'm subtracting 7<em>y</em>, though I don't like dealing with negatives.

7<em>y</em> – 7<em>y</em> – 36 = <em>y</em> – 7<em>y</em>  — subtract 7<em>y</em> from both sides

–36 = –6<em>y</em>  — simplify

–36 ÷ –6 = –6<em>y</em> ÷ –6  — divide by –6 on both sides

<em>y</em> = 6  — simplify

Again, we can diverge here: substitute <em>y</em> into 3<em>x</em> – 30 = <em>y</em> or substitute <em>y</em> into 7<em>y</em> – 6 = 3<em>x</em>

I'm going to choose 3<em>x</em> – 30 = <em>y</em> but it will work either way, should you take the time (if you have it) to chase down every path this problem can take.

3<em>x</em> – 30 = <em>y</em>  — initial equation

3<em>x</em> – 30 = 6  — substitute 6 for <em>y</em>

3<em>x</em> – 30 + 30 = 6 + 30  — add 30 to both sides to isolate 3<em>x</em>

3<em>x</em> = 36  — simplify the expression

3<em>x</em> ÷ 3 = 36 ÷ 3  — divide both sides by 3 to isolate <em>x</em>

<em>x</em> = 12  — simplify

So, we have <em>x</em> = 12 and <em>y</em> = 6 . We know they work for 3<em>x</em> – 30 = <em>y</em>  but not if they work for 7<em>y</em> – 6 = 3<em>x</em> . Let's substitute those in to see if (12, 6) really is the solution point.

7<em>y</em> – 6 = 3<em>x</em>  — original equation

7(6) – 6 ≟ 3(12)  — substitute 6 for <em>y</em> and 12 for <em>x</em>

42 – 6 ≟ 36  — simplify by multiplying

36 = 36 ✔  — simplify by combining like terms on left side

Success! It works! We have found our solution!

I hope this helps increase your understanding of the concept. Have a great day!

8 0
3 years ago
PLEASE ANSWER ILL GIVE 5 STARS AND A HEART!!!
r-ruslan [8.4K]

Answer:

120

Step-by-step explanation:

5x6 = 30

30x4= 120

7 0
2 years ago
Let a, b, c, and d be non-zero real numbers. If the quadratic equation ax(cx + d) = -b(cx+d) is solved for x, which of the follo
LiRa [457]

Answer:

Option d

Step-by-step explanation:

given that a, b, c, and d be non-zero real numbers.

ax(cx + d) = -b(cx+d) \\acx^2+x(ad+bc)+bd =0

we can factorise this equation by grouping

(acx^2+xad)+)xbc+bd) =0\\ax(cx+d) +b(cx+d) =0\\(ax+b)(cx+d) =0

Equate each factor to 0 to get

x=\frac{-b}{a} , \frac{-d}{c}

Ratio of one solution to another would be

\frac{-b}{a} / \frac{-d}{c} \\=\frac{ad}{bc}

So ratio would be ad/bc

Out of the four options given, option d is equal to this

So option d is right

4 0
3 years ago
A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the
amm1812

Answer:

s(t)=(ln|7|+ln|2|)\,ft

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\

Integrating both sides w.r.to t

v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\

Using initial value

v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}

We have to calculate the distance covered in time interval [0,5], so:

\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft

3 0
3 years ago
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