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Zepler [3.9K]
3 years ago
8

What is log base 5(4*7 )+log base 5 of 2 written as a single log?

Mathematics
2 answers:
nataly862011 [7]3 years ago
7 0
The answer for the exercise shown above is the last option (Option D), which is:

 D. log base 5 of 56

The explanation is show below:

 1. You have the following logarithm expresssion:

 <span>log5(4*7 )+log5(2)
</span>
 2. By the logarithms properties, you can rewrite the logarithm expression as following:

 log5(28)(2)
 log5(56)

 3. Therefore, as you can see, the answer is the option mention before.
 
sergey [27]3 years ago
3 0

Answer:

D. log base 5 of 56

Step-by-step explanation:

Since, by the multiplication law of logarithm,

log_a(m)+log_a(n) = log_a(mn)

Given expression is,

log base 5(4*7 )+log base 5 of 2

=log_5(4\times 7)+log_5(2)

=log_5(28)+log_5(2)

By the above property,

=log_5(28\times 2)

=log_5(56)

= log base 5 of 56

Hence, Option D is correct.

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3 years ago
Jim picks a number he doubles that number then adds five and gets 41 as his result what was the original number Jim picked
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Work out the value of (2^3)^2
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Step-by-step explanation:

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2 years ago
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The triangular prism below has a height of 9 units and a volume of 162 units. Find
DanielleElmas [232]

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Step-by-step explanation:

Using the Triangular Prism formula,

V=A (sub)B(sub) h

Solving for A (sub)B(sub)

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Hope this helps...

6 0
2 years ago
The coordinates of three of the vertices of parallelogram ABCD are A(1, 0), B(2, 3), C(3, 2). What are coordinates of the fourth
nalin [4]

Answer:

 A) The coordinates of the fourth vertex are:

1) x-coordinate:

x=2

2) y-coordinate:

y=-1

B) The point of intersection of the diagonals is:  (2,1)

Step-by-step explanation:

We need to remember that the diagonals of a parallelogram intersect each other at a half-way point and the midpoint of each diagonal is the same.

The midpoint formula is:

M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Since:

M_{AC}=M_{BD}

We can find the coordinates of the fourth vertex D(x,y) through these procedure:

1) x-coordinate:

\frac{1+3}{2}=\frac{2+x}{2}\\\\2(2)=x\\\\4-2=x\\\\x=2

2) y-coordinate:

\frac{0+2}{2}=\frac{3+y}{2}\\\\1(2)-3=y\\\\y=-1

Therefore,  fourth vertex is D(2,-1)

Since the point of intersection of the diagonals is the midpoint of a diagonal (Remember that  M_{AC}=M_{BD}), this is:

M_{AC}=M_{BD}=(\frac{1+3}{2},\frac{0+2}{2})\\\\M_{AC}=M_{BD}=(2,1)

Therefore, the point of intersection of the diagonals is (2,1)

7 0
3 years ago
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