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sertanlavr [38]
2 years ago
12

5y-4x=-7 what does x equal

Mathematics
1 answer:
harkovskaia [24]2 years ago
4 0

Answer:

x = \frac{5y + 7}{4}

Step-by-step explanation:

5y - 4x = - 7

5y - 4x + 4x = - 7 + 4x         [adding + 4x on both sides ]

5y + 7 = - 7 + 7 + 4x            [adding 7 on both sides , - 4x + 4x = 0 ]

5y + 7 = 4x                         [ -7 + 7 = 0 ]

\frac{5y + 7}{4}  = \frac{4x}{4}                            [ dividing by 4 on both sides ]

x = \frac{5y + 7}{4}

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I have a test on these type of questions and don't know what to do....
3241004551 [841]

Answer:

You take the info and answer the questions with guidance from the tables

Step-by-step explanation:

4 0
2 years ago
I rent a gym for $150 for 30 students another time I rent the gym for $270 for 70 students I need to find a fixed rate
kirza4 [7]

Answer:

Fixed rate is $60.

Step-by-step explanation:

Let us consider per student charge be 'x'.

Let us consider fixed rate be 'b'

Given:

I rent a gym for $150 for 30 students.

So we can say that;

Total amount is equal to sum of number of students multiplied by per student charge and fixed rate.

framing in equation form we get;

30x+b=150 \ \ \ \ equation \ 1

Also Given:

another time I rent the gym for $270 for 70 students.

So we can say that;

Total amount is equal to sum of number of students multiplied by per student charge and fixed rate.

framing in equation form we get;

70x+b=270 \ \ \ \ equation \ 2

Now we will subtract equation 1 from equation 2 we get;

70x+b-(30x+b)=270-150\\\\70x+b-30x-b=120\\\\40x=120

Dividing both side by 40 we get;

\frac{40x}{40}=\frac{120}{40}\\\\x=\$3

Now we will substitute the value of x in equation 1 we get;

30x+b=150\\\\30\times3+b=150\\\\90+b=150\\\\b=150-90 =\$60

So we can say that the equation can be written as;

y =3x+60

Hence we can say that fixed rate is $60 and per student charge is $3.

3 0
3 years ago
Read 2 more answers
12 more than 8.2 times a number n
snow_tiger [21]

Answer:

12+8.2n, this is the answer

8 0
3 years ago
You can choose between two telephone companies. Company A has a monthly charge of $10 plus 5¢ a minute for each minute of long d
Diano4ka-milaya [45]

Answer: Company A would be a better deal for number of minutes over 200

Step-by-step explanation:

Let x represent the number of minutes that you would use with either company A or company B.

Company A has a monthly charge of $10 plus 5¢ a minute for each minute of long distance calls. It means that the cost of x minutes of long distance call is

10 + 0.05x

Company B has a monthly charge of $8 plus 6¢ a minute for long distance calls. It means that the cost of x minutes of long distance call is 10 + 0.05x

8 + 0.06x

For Company A to be a better deal, it means that

10 + 0.05x < 8 + 0.06x

10 - 8 < 0.06x - 0.05x

2 < 0.01x

0.01x > 2

x > 2/0.01

x > 200

5 0
3 years ago
Find a and b such that gcd(a, b) = 14, a &gt; 2000, b &gt; 2000, and the only prime divisors of a and b are 2 and 7.
choli [55]

Answer:

a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3

Step-by-step explanation:

As 2  and 7 are the only prime divisors of both a and b we know that both can be written as:

a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\

Where \mathbb{N}_{0} is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).

We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:

a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\

Why do I write them like this? Because this way is easier to observe that if \alpha and \gamma were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if \beta and \delta were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.

So either of them has to be equal to zero. And then we have that

a = 2^\alpha.14 \\b = 7^\delta.14

All we have left to do is find the possible values for \alpha and \delta so that a>2000 \ , b>2000 and that only happens if \alpha\geq 8 and \delta\geq 3

8 0
3 years ago
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