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nika2105 [10]
3 years ago
7

Pablo wants to save $800 to buy a TV. He saves $19 each week. The amount, A (in dollars), that he still needs after

Mathematics
2 answers:
Aleks04 [339]3 years ago
8 0

Answer:

a) $686

b) 17 weeks

Step-by-step explanation:

a) Set w=6, as Pablo has already saved for 6 weeks. Plug in the value and solve for A(6).

A(w)=800-19w

A(6)=800-19(6)

A(6)=800-114

A(6)=686

Pablo still needs $686.

b) Set A(w)=477, as it is the "final amount" for that week.  Solve for w.

A(w)=800-19w

477=800-19w

-323=-19w

w=17

Pablo has been saving for 17 weeks if he still needs $477.

Sidana [21]3 years ago
6 0

Answer:

Pablo wants to save $800 to buy a TV.

He saves $19 each week.

The amount, A (in dollars), that he still needs after  weeks is given by the following function.

A(w)= 800-19w

(a) How much money does Pablo still need after 6 weeks?

Putting w = 6 in the equation;

A(6)= 800-19(6)

= $686

(b) If Pablo still needs $477, how many weeks has he been saving?

This means Pablo has already saved 800-477=323 dollars

Means he has saved for \frac{323}{19}=17 weeks.

He has been saving for 17 weeks now.

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Now we know if we subtract 3 from the width, we have a new rectangle that gives us an area of 24 inches. Therefore, our new triangle has the following:

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Answer:

2*log(x)+log(y)

Step-by-step explanation:

So, there are two logarithmic identities you're going to need to know.

<em>Logarithm of a power</em>:

   log_ba^c=c*log_ba

   So to provide a quick proof and intuition as to why this works, let's consider the following logarithm: log_ba=x\implies b^x=a

   Now if we raise both sides to the power of c, we get the following equation: (b^x)^c=a^c

   Using the exponential identity: (x^a)^c=x^{a*c}

    We get the equation: b^{xc}=a^c

    If we convert this back into logarithmic form we get: log_ba^c=x*c

    Since x was the basic logarithm we started with, we substitute it back in, to get the equation: log_ba^c=c*log_ba

Now the second logarithmic property you need to know is

<em>The Logarithm of a Product</em>:

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    Now for a quick proof, let's just say: x=log_ba\text{ and }y=log_bc

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   b^{x+y}=ac

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