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Lilit [14]
3 years ago
5

4. Un cuadrado tiene un área de 400 cm2; si se aumenta un centímetro por lado, ¿cuál será el perímetro del nuevo cuadrado?

Mathematics
1 answer:
CaHeK987 [17]3 years ago
6 0

Step-by-step explanation:

I can't read Spanish or read Spanish I'm sorry

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the production line where you work can assemble 5 amplifiers every 30 minutes at this rate how long should it take the line to a
nevsk [136]
See this question as Ratio.

Amplifiers  :   Time Taken in minutes
    
      5          :          30
   
     5x= 125
       x= 25

       so

5*25 : 30*25

125 : 750


750 minutes.
 12 hours 30 minutes.

6 0
3 years ago
Pls help me with this question!!!
yarga [219]

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20% off the regular price of $32.50
yuradex [85]
You multiply the original price by .20
7 0
3 years ago
6x+3x-(x-10)=20x help... step by step.
mihalych1998 [28]
In equations we always want to do the same thing to both sides so the equation stays equal

in an equation with one unknown (x) we try to get the unknown on one side

so -(x-10)
this means that you multiply everything in the equation by -1 or in other words you make everything in the parenthasees the opposite sign of what it is so

6x+3x-x+10=20x
add like terms
8x+10=20x
subtract 8x from boths sides
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6 0
4 years ago
Read 2 more answers
PLEASE HELP!!
Bezzdna [24]

Answer:

y=-\dfrac{9}{40}x^2+\dfrac{441}{40}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}

Step-by-step explanation:

If a parabola has its vertex on the y-axis, then its equation is

y=ax^2+b

This parabola passes through the point R(3,9), then

9=a\cdot 3^2+b\\ \\9=9a+b

The area of the right triangle PQR is

A_{PQR}=\dfrac{1}{2}\cdot PQ\cdot QR

Find PQ and QR, if P(x_1,0),\ Q(3,0),\ R(3,9):

PQ=\sqrt{(x_1-3)^2+(0-0)^2}=|x_1-3|\\\\QR=\sqrt{(3-3)^2+(0-9)^2}=9

Now,

40.5=\dfrac{1}{2}\cdot |x_1-3|\cdot 9\\ \\90=9|x_1-3|\\ \\|x_1-3|=10\\ \\x_1-3=10\ \text{or}\ x_1-3=-10\\ \\x_1=13\ \text{or }x_1=-7

We get two possible points P_1(-7,0) and P_2(13,0).

For point P_1:\\

0=a\cdot (-7)^2+b\\ \\49a+b=0

So,

b=-49a\\ \\9=9a-49a\\ \\-40a=9\\ \\a=-\dfrac{9}{40}\\ \\b=\dfrac{441}{40}\\ \\y=-\dfrac{9}{40}x^2+\dfrac{441}{40}

For point P_2:\\

0=a\cdot (13)^2+b\\ \\169a+b=0

So,

b=-169a\\ \\9=9a-169a\\ \\-160a=9\\ \\a=-\dfrac{9}{160}\\ \\b=\dfrac{1,521}{160}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}

5 0
3 years ago
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