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Arte-miy333 [17]
3 years ago
15

Determine if the following equations are equal by evaluating both sides

Mathematics
1 answer:
padilas [110]3 years ago
5 0

Answer:

Yes, the both sides of the given equation are equal.

Step-by-step explanation:

The given equation is

\log((1-i)^3)=3\log(1-i)

Taking LHS,

LHS=\log((1-i)^3)

Using the power property of logarithm, we get

LHS=3\log(1-i)                               [\because log_ax^n=nlog_ax]

LHS=RHS                                   [\because RHS=3\log(1-i)]

Both sides of the given equation are equal.

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Find vector c if c=a+b (view image)<br> A. 7,6<br> B. -7,2<br> C. 0,-5<br> D. -5, 0
o-na [289]

Answer:

\vec c=

Option B

Step-by-step explanation:

<u>Sum of vectors in</u> \mathbb{R}^2

Given two vectors \vec a= and \vec b=. The sum of both vectors is:

\vec a+\vec b =

The vectors in the figure are

\vec a=

\vec b=

The sum is

\vec c=\vec a+\vec b =

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Option B

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2 years ago
A music store bought a CD set at a cost of $20. When the store sold the CD set, the percent markup was 35%. Find the selling pri
lbvjy [14]

Answer:

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Step-by-step explanation:

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3 years ago
Kaitlynn needs 3 pounds of fruit for a salad.
gizmo_the_mogwai [7]
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2 years ago
(2) Last month, Coach Harvey ran 48 miles. This was 60% of his goal. How many miles did Coach Harvey set as his goal? (Set a pro
ivolga24 [154]

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2 years ago
Find the distance between the two points in simplest radical form.
Jlenok [28]

Answer:

d=\sqrt{58}\approx7.6

Step-by-step explanation:

To find the distance between any two points, we can use the distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Our first point, A, is at (1, 1) and our second point, B, is at (-2, 8).

Let's let A(1, 1) be (x₁, y₁) and B(-2, 8) be (x₂, y₂). Substitute this into the distance formula:

d=\sqrt{(-2-1)^2+(8-1)^2

Subtract:

d=\sqrt{(-3)^2+(7)^2

Square:

d=\sqrt{9+49}

Add:

d=\sqrt{58}

This cannot be simplified.

So, the distance between the two points is √58 or about 7.6 units.

And we're done!

So... where's my cookie :)?

6 0
3 years ago
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