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madam [21]
3 years ago
13

For what value of q is in the question -13(q+4)=7q+16​

Mathematics
1 answer:
Rufina [12.5K]3 years ago
5 0

Answer:

q=-3 2/5

Step-by-step explanation:

First multiply -13 with (q+4)

-13q-52=7q+16

Now you have to change the order before = the numbers with q and after = the numbers without anything

-13q-7q=52+16

Remember that when you change positions the symbol too

Then solve it

-20q=68

Now u have to divide as i do

q=68/-20

you can simplify it

q=3 8/-20

Now divide by the greater divisor in this case is 4

q=3 2/-5

And organize it

q=-3 2/5

Never forget to do q=(answer)

Hopefully it helps.

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there are nine single discs in a classroom the rest of the deaths are double desks the classroom has seating for 33 learners how
mixas84 [53]
9+2x=33
2x=24
X=12 double desks

8 0
3 years ago
A mechanic used 49 quarts of oil in 7 cars. At this rate, how much oil would he use to change the oil in 17 cars?
Karo-lina-s [1.5K]

Answer:

Each car used 7 quarts of oil. So 7 cars would use 7×17 quarts which would be 119 quarts.

8 0
3 years ago
Jacob is a plumber. Due to increased material costs, he puts all his prices up by 7%. He now chargers R320 per hour for her serv
Lorico [155]

Answer:

R290.1

Step-by-step explanation:

Let

x = charge before he put his prices up

New charge = R320

Percentage increase in price = 7%

x + 7% of x = 320

x + 0.07 * x = 320

x + 0.07x = 320

1.07x = 320

x = 320/1.07

x = 299.06542056074

Approximately

x = R290.1

Charge before he put his prices up = R290.1

4 0
3 years ago
50 men working 10 hours daily can complete a job in
Sliva [168]

<h2>Given :-</h2>

  • 50 men do 10 hours work daily complete work in 20 days

  • 80 men do job in 50% time

<h2>To Find :-</h2>

  • Days required

<h2>Solution :-</h2>

  • 1 man may do job in (50 × 20) days =
  • 1 man may do job in 1000 days

Now,

  • 80 man = 1000/80 = 12.5

Now,

  • When hours reduced 50% then we will multiply days by 2
  • Total days = 12.5 × 2 = 25 days.

<h2>Hence</h2>

  • Option C is correct

\begin{gathered} \\ \end{gathered}

8 0
3 years ago
When the solutions are going to be complex, which methods can be used to solve quadratic equations?
yKpoI14uk [10]
You could complete the square to state the vertex. 
You could use the quadratic equation to find the roots (which are complex). 

Try an example that will require both.

y = x^2 + 2x + 5

Step One
Get the graph. That's included below.

Step Two
Provide the steps for completing the square.
Note: we should get (-1,4)

y= (x^2 +2x ) + 5
y = (x^2 +2x + 1) + 5 - 1
y = (x +1)^2 + 4

The vertex is at (-1,4)

Step Three
Find the roots. Use the quadratic equation. Note that the graph shows us that the equation never crosses or touches the x axis. The roots are complex.

\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}

a = 1
b = 2
c = 5

\text{x = }\dfrac{ -2 \pm \sqrt{2^{2} - 4*1*5 } }{2}
\text{x = }\dfrac{ -2 \pm \sqrt{4 - 20 } }{2}
\text{x = }\dfrac{ -2 \pm \sqrt{-16 } }{2}
\text{x = }\dfrac{ -2 \pm 4i }{2}
x = -1 +/- 2i

x1 = -1 + 2i
x2 = -1 - 2i And we are done.


7 0
3 years ago
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