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3 years ago

12

How can the properties of exponents help solve exponential equations?

1 answer:

qwelly [4]3 years ago

7 0

The properties of exponents would help solve exponential equations by creating an expression that can be simplified easily. It elps in simplfying a certain equation. It tells us how are terms that contain exponents can be added, subtracted, divided and multiplied. For addition, terms having the same can only be added. This would be also true for subtraction. For example, we have x^2 + y^3, these terms could not be added since they have different base. For multiplication, we can readily multiply these terms and add the exponents of the terms with the same base, x^2(y^2)(x) = x^3y^2. For division, they can also be divided readily but, terms with the same base, the exponents are to be subtracted, x^3 / x^2 = x.

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XYZ Corporation recorded the number of employee absences each week over a period of 10 weeks. The result is the data list 5, 3,

nadezda [96]

Answer:

4 ansences

Step-by-step explanation:

40 / 4 is 10 so ans is 10 absences

8 0

3 years ago

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and increases from quadrant 2 into quadrant 1. It

dusya [7]

Answer:

f(x) = $2^{(3x)}$

Step-by-step explanation:

The given options are,

a) f(x) = $2^{(3x)}$

b) f(x) = $({\frac {1}{2}})^{3x}$

c) f(x) = $2 \times ({\frac {1}{3}})^{x}$

d) f(x) = $\frac {1}{2} \times ({\frac {1}{3}})^{x}$

Now, among the given options only option (a) satisfies the criteria stated in the question i. e.,

1. f(x) → 0 as x → - ∞ and

2 f(0) = 1

so, the correct answer is,

f(x) = $2^{(3x)}$

7 0

4 years ago

Read 2 more answers

How do you break apart the factor 42 using place values

VLD [36.1K]

Let's break apart 42.

The 4 is in the tens. The 2 is in the ones.

Place Value starting from Thousands

Thousands Hundreds Tens Ones

_________ ________ ____ _____

4 2

So this is how we break up numbers using place value.

Hope this helps!

5 0

3 years ago

A = √7 + √c and b = √63 + √d where c and d are positive integers.

mrs_skeptik [129]

Answer:

$\displaystyle a:b=\frac{1}{3}$

Step-by-step explanation:

<u>Ratios </u>

We are given the following relations:

$a=\sqrt{7}+\sqrt{c}\qquad \qquad[1]$

$b=\sqrt{63}+\sqrt{d}\qquad \qquad[2]$

$\displaystyle \frac{c}{d}=\frac{1}{9} \qquad \qquad [3]$

From [3]:

$9c=d$

Replacing into [2]:

$b=\sqrt{63}+\sqrt{9c}$

We can express 63=9*7:

$b=\sqrt{9*7}+\sqrt{9c}$

Taking the square root of 9:

$b=3\sqrt{7}+3\sqrt{c}$

Factoring:

$b=3(\sqrt{7}+\sqrt{c})$

Find the ration a:b:

$\displaystyle a:b=\frac{\sqrt{7}+\sqrt{c}}{3(\sqrt{7}+\sqrt{c})}$

Simplifying:

$\boxed{a:b=\frac{1}{3}}$

3 0

3 years ago

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Lelu [443]

Answer:

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Step-by-step explanation:

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3 years ago