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Vikentia [17]
3 years ago
6

100 points if you answer all these questions

Mathematics
2 answers:
Aleksandr [31]3 years ago
5 0

Answers:

<h3>Distributive Property</h3><h3>Addition Property of Equality</h3><h3>Division Property of Equality</h3>

Step-by-step explanation:

Let's solve this problem step-by step to see what we use to solve this.

<h3>Step 1: Distributive Property</h3>

3x - 9 = 12

<h3>Step 2: Addition Property of Equality</h3>

3x = 21

<h3>Step 3: Division Property of Equality</h3>

x = 7

I'm always happy to help :)

almond37 [142]3 years ago
3 0

Distributive Property, Addition Property of Equality, Division Property of Equality

To solve this problem is to see what we have to  use to solve this.

First step: Distributive Property: 3x - 9 = 12

Second step: Addition Property of Equality: 3x = 21

Third step: Division Property of Equality: x = 7

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Vladimir [108]
5+-2x=2x+-5
5=4x+-5 (add 2x to both sides)
10=4x (add 5 to both sides)
10/4=x (divide both sides by 4)
3 0
3 years ago
What is the domain of the function shown in the graph below?
cricket20 [7]

Answer: interval from -8 to 5

In interval notation, we would write [-8, 5]

The domain is the set of allowed x inputs to a function. The left-most point is (-8,0) so x = -8 is the smallest x value allowed. The largest x value allowed is x = 5 due to the point (5,0) being the right-most point.

The use of square brackets in interval notation says "include this endpoint as part of the interval".

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3 years ago
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Step-by-step explanation:

6 0
3 years ago
If log2 5 = k, determine an expression for log32 5 in terms of k.
lukranit [14]

Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

6 0
3 years ago
Its add math. i really need help, solve the question. thanks♥️​
dexar [7]

Answer:

see explanation

Step-by-step explanation:

Given

f(x) = ax + b

x = 2 → f(x) = 1 and x = - 1 → f(x) = - 5 ( from the table )

Substitute these values into f(x) = ax + b, that is

2a + b = 1 → (1)

- a + b = - 5 → (2)

Subtract (2) from (1) term by term to eliminate b

3a = 6 ( divide both sides by 3 )

a = 2

Substitute a = 2 into (2) and evaluate for b

- 2 + b = - 5 ( add 2 to both sides )

b = - 3

(b)

When x maps onto itself then

ax + b = x, that is

2x - 3 = x ( subtract x from both sides )

x - 3 = 0 ( add 3 to both sides )

x = 3

Thus a = 2, b = - 3 and x = 3

6 0
3 years ago
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