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MrRa [10]
3 years ago
13

Mr. Lopez has 238 silver coins in his coin collection. The actual weight of the real silver in each coin is 0.85 ounce. How many

ounces of real silver does Mr. Lopez have in his coin collection?
Mathematics
2 answers:
mars1129 [50]3 years ago
6 0

Answer:

208?

Step-by-step explanation:

im not sure if this is 100% correct but, all I did was:

238/0.85

=208

NNADVOKAT [17]3 years ago
3 0

Answer:

202.3 ounces of real silver

Step-by-step explanation:

Another example would be like pizza. If there are 6 slices of pizza in each box, then 2 boxes of pizza would have 6 + 6 slices = 12 slices. It is 6 slices each * 2 boxes = 12 slices total. The same applies to the coin situation, for every additional coin, multiply 0.85 by the number of coins to get the total weight of real silver.

If there are 238 silver coins, then you have 238 0.85 ounces of real silver. Multiplying 238 by 0.85 to simplify gives you 202.3 ounces of real silver.

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Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
3 years ago
Find the 11th term of the following geometric sequence.<br> 1, 3, 9, 27, ...
Natasha_Volkova [10]

Answer:

the answer is 59048

Step-by-step explanation:

as the common ratio is multiplying by 3 and the first term is 1 so from the rule (Tn=ar(power n-1 ) )

so the 11th term is 1*3(power 11-1 ) equal 3 power 10 equal 59048

4 0
3 years ago
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3 years ago
Which line must be perpendicular to ​
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3 years ago
Read 2 more answers
Find the formula for an exponential function that passes through the two points given.
goldenfox [79]

Answer:

y = 7(3^x)

Step-by-step explanation:

Given

(x_1,y_1) = (0,7)

(x_2,y_2) = (4,567)

Required

Determine the formula

An exponential function is of the form:

y = ab^x

For point (x_1,y_1) = (0,7)

7 = ab^0

7 = a*1

7 = a

a = 7

For point (x_2,y_2) = (4,567)

567 = ab^4

Substitute 7 for a

567 = 7*b^4

Divide both sides by 7

81 = b^4

Take 4th root of both sides

\sqrt[4]{81} =\sqrt[4]{b^4}

\sqrt[4]{81} =b

b = \sqrt[4]{81}

b = 3

Substitute 7 for a and 3 for b in y = ab^x

y = 7 * 3^x

y = 7(3^x)

3 0
3 years ago
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