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avanturin [10]
2 years ago
10

A Leopoldo su Maestra l le pidió que calcular el perímetro de un rectángulo que tiene las medidas que se indican en el dibujo cu

ánto mide su perímetro
Mathematics
1 answer:
guapka [62]2 years ago
7 0

Answer:

Lo siento, no puedo ayudarte, porque esta resonancia es porque no tienes una imagen ni nada para que yo te ayude, lo siento, espero que tengas un buen día.

Step-by-step explanation:

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7-5=2 so what is 7-2=? , free
Sergio [31]

Answer:

If 7 minus 5 equals 2, then 7 minus 2 equals 5.

Step-by-step explanation:

This is because of the associative property of equality.

<em><u>Please post #teamtrees #PAW (Plant And Water) if you have social media :)</u></em>

6 0
2 years ago
Hillel is juggling flaming torches to raise money for charity. His initial appearance raises $500, and he raises $15, for each m
avanturin [10]
15=1
so
R = t*15

(Pretty sure)
4 0
3 years ago
Read 2 more answers
Which ordered pair is a solution of the equation?<br> Y-4=7(x-6)
deff fn [24]

Answer:

(6,4)

Step-by-step explanation:

The equation is in 'point-slope' form.

y-y_1=m(x-x_1)\\\rule{150}{0.5}\\(x_1,y_1) \rightarrow \text{a point that's given}\\\\m \rightarrow \text{slope}

This means we can identify the point that was used in the equation.

y-\boxed{y_1} =m(x-\boxed{x_1}) | y-\boxed{4}=7(x-\boxed{6})\\\rule{210}{0.5}\\y_1 = 4\\\\x_1 =6\\\rule{210}{0.5}\\(x_1,y_1) \rightarrow \boxed{(6,4)}

(6,4) would be a solution to the equation given.

Hope this helps.

4 0
2 years ago
What is the true solution to 2 l n e Superscript l n 5 x Baseline = 2 l n 15 x = 0 x = 3 x = 9 x = 15
weeeeeb [17]

Answer:

x = 3

Step-by-step explanation:

I got it right on edge

4 0
3 years ago
In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
WARRIOR [948]

Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.

3 0
3 years ago
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