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Sauron [17]
3 years ago
11

During a store's closing sale, everything is 65% off. Vikram goes into the store to buy a shirt and a pair of pants, and spends

15.60 before tax. How much would he have spent without the sale?
Mathematics
1 answer:
saw5 [17]3 years ago
5 0

Answer:

without the sales, he would have spent $44.57

Step-by-step explanation:

Let the amount he would have spent without the sales be x. Now, if there is a 65% discount, what this means is that he is exactly paying for 100 - 65% = 35%

Now, it is this 35% of X that is equal to the amount he paid

Thus, mathematically, we have the following;

35/100 * x = 15.60

35x = 15.6 * 100

35x = 1560

x = 1560/35

x = $44.57

This means that without the sales discount, he would have paid $44.57

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Help ASAP show work please thanksss!!!!
Llana [10]

Answer:

\displaystyle log_\frac{1}{2}(64)=-6

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

log_b(1) = 0

Logarithm of the base:

log_b(b) = 1

Product rule:

log_b(xy) = log_b(x) + log_b(y)

Division rule:

\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)

Power rule:

log_b(x^n) = n\cdot log_b(x)

Change of base:

\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

\displaystyle log_\frac{1}{2}(64)

Factoring 64=2^6.

\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)

Since

\displaystyle 2=(1/2)^{-1}

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})

Applying the logarithm of the base:

\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}

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2 years ago
In 1960, the population of a town was 14 thousand people. Over the course of the next 50 years, the town grew at a rate of 30 pe
andre [41]

SOLUTIONS

Given: Assume y is the population

\begin{gathered} t=0=1960,y=14030 \\ t=1=1961,y=14060 \\ so\text{ b = 14000} \\ so\text{ k = 30} \end{gathered}

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\begin{gathered} 2030-1960=70 \\ y=30k+14000 \\ y=30(70)+14000 \\ y=2100+14000 \\ y=16100people \end{gathered}

(B) so when y = 15000, find t

\begin{gathered} y=15000 \\ y=30t+14000 \\ 15000=30t+14000 \\ 15000-14000=30t \\ 1000=30t \\ t=\frac{1000}{30} \\ t\approx33 \\ t=1960+33 \\ t=1993 \end{gathered}

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