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3 years ago

What is the answer to this?

Mathematics

2 answers:

Zepler [3.9K]3 years ago

8 0

Answer:

67 units

Step-by-step explanation:

Shkiper50 [21]3 years ago

6 0

Answer:

67 units squared

Step-by-step explanation:

First, you multiply all the combinations possible with the 3 values.

5x1.5, 5x4, 1.5x4

But there are 2 of every side so you multiply each by 2.

5x1.5x2 = 15

5x4x2 = 40

1.5x4x2 = 12

12+40+15 = 67

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Radious is given by the function c(R)=5.25r+125

Mamont248 [21]

Answer:

pie eguals 3.14 do it your self

Step-by-step explanation:

8 0

2 years ago

Does 3 work as a solution to the following inequality? 2n > 6

Ksju [112]

Answer:

No.

Step-by-step explanation:

If n=3 the only thing you get is 6, which is not more than 6.

5 0

2 years ago

if you shop in chicago, Illinois, the state sales tax is 9 1/4% what is your total cost when you purchase a $5.00 map of the cit

NNADVOKAT [17]

1.) 1 ÷ 4 =0.25 + 9= 9.25

2.) 9.25 + 5.00 = 14.25

7 0

3 years ago

Please show your work.

Nastasia [14]

Answer:

1. 8.19, 2. 25.96, 3. 3.95

Step-by-step explanation:

use soh cah toa

1. cos is adjacent/hypotenuse

cos73=x/28

multiply both sides by 28

28cos73=x=8.186

2. sin is opposite over hypotenuse

sin68=x/28

multiply both sides by 28

28sin68=x=25.961

3. sin is opposite over hypotenuse

sin26=x/9

myltiply both sides by 9

9sin26=x=3.945

8 0

2 years ago

HELLOOOO HELP PLEASE

MA_775_DIABLO [31]

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>:

$log_b{ac}=log_ba+log_bc$

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

Now rewriting them both in exponential form, we get the equations:

$b^x=a\\b^y=c$

We can multiply a * c, and since b^x = a, and b^y = c, we can substitute that in for a * c, to get the following equation:

$b^x*b^y=a*c$

Using the exponential identity: $x^{a}*x^b=x^{a+b}$ , we can rewrite the equation as:

$b^{x+y}=ac$

taking the logarithm of both sides, we get:

$log_bac=x+y$

Since x and y are just the logarithms we started with, we can substitute them back in to get: $log_bac=log_ba+log_bc$

Now let's use these identities to rewrite the equation you gave

$log(x^2y)$

As you can see, this is a log of products, so we can separate it into two logarithms (with the same base)

$log(x^2)+log(y)$

Now using the logarithm of a power to rewrite the log(x^2) we get:

$2*log(x)+log(y)$

3 0

1 year ago