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Airida [17]
3 years ago
10

During cooking, chicken loses 10% of its weight due to water loss. In order to obtain 1,170 grams of cooked chicken, how many gr

ams of uncooked chicken must be used?
Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
5 0

Answer:

117 grams

Step-by-step explanation:

So this is a oddly phrased question but I believe I understand it. You need to find 10 percent of 1170 grams and that it 117. And for you information I did use calculator to prevent error so don't worry about that.

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Un padre tiene actualmente cuatro veces la edad de su hijo. Cuando pasen 5 años, su edad será solo tres veces superior. ¿Qué eda
julsineya [31]

Answer:

I. Hijo, h = 10 años

II. Padre, P = 40 años

Step-by-step explanation:

  • Sea la edad del padre P.
  • Sea la edad del hijo h.

Traduciendo el problema verbal a una expresión algebraica, tenemos;

P = 4h .....ecuación 1.

P + 5 = 3(h + 5) ........ecuación 2.

Simplificando aún más, tenemos;

P + 5 = 3h + 15

P = 3h + 15 - 5

F = 3h + 10 ......ecuación 3.

Sustituyendo la ecuación 1 en la ecuación 3;

4h = 3h + 10

4h - 3h = 10

<em>h = 10 años </em>

A continuación, encontraríamos la edad del padre;

P = 4h

P = 4 * 10

<em>P = 40 años</em>

4 0
3 years ago
You and your friend are selling magazine subscriptions for a fundraiser. After w weeks, you have sold (7w+6) subscriptions and y
Mariulka [41]
You have sold 72, and your friend has sold 74. Meaning your friend has sold 2 more magazine subscriptions than you.
5 0
3 years ago
Find (f^-1)(a). F(x)=cosx- sin x + pi/2 a=pi/2, 0
Novosadov [1.4K]

Answer:

pi/4

Step-by-step explanation:

my answer is in the image above

7 0
3 years ago
470.47 in expanded form​
Aleksandr-060686 [28]

Four hundred seventy and forty seven hundredth

7 0
4 years ago
Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0
3 years ago
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