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son4ous [18]
3 years ago
5

Two models of cellular telephones, red and blue, are stored in boxes. One box weighs twelve pounds and contains four of the red

models and one blue model. Another box weighs eight pounds and contains one blue model and two red models. How much does each model weigh?
Mathematics
1 answer:
GenaCL600 [577]3 years ago
5 0
First, make a equation in which r= red and b=blue.

So, since in the first box it has 4 red models and one blue model equaling 12,

the first equation looks like 4r+b=12.

The second equation looks like 2r+b=8.

What you would do is try and first solve for r by getting rid of b.

Since both equation has a positive b, you would make one equation have a negative b by multiplying the whole equation by -1.

-1*(2r+b=8)= -2r-b=-8

Add.

4r+b=12
+(-2r-b=-8)

2r=4
r=2

Then, you plug in 2 for the r for any of the original equations.

4(2)+b=12

8+b=12

b=4

or

2(2)+b=8

4+b=8
b=4

So, the red models weigh 2 pounds while the blue models weigh 4.



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3 years ago
What is the rate of change between (29,9) and (33,10)?
Tema [17]

Answer:

\frac{1}{4}

Step-by-step explanation:

The rate of change is a measure of the slope

Calculate the slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (29, 9) and (x₂, y₂ ) = (33, 10)

m = \frac{10-9}{33-29} = \frac{1}{4} ← rate of change

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3 years ago
Helppppppppppp plssss
Reptile [31]

Answer:

It is the 3rd choice. It is 4 times as big as the smaller cylinder

Step-by-step explanation:

V = πr2h

r - radius

h - height

π - pi

  • Small cyclinder

= π(3)2(10)

enter into a calucaltor and you get 282.74

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= π(6)2(10)

enter into a calucaltor and you get 1130.97

Divide 1130 by 282 and you get about 4

3 0
3 years ago
I will mark as brainliest if you can answer. :)
Yuri [45]

not enough information

8 0
3 years ago
If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(
timama [110]

Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

to write

xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

to write

xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
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