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grigory [225]
3 years ago
13

I need help with this ?

Mathematics
1 answer:
BartSMP [9]3 years ago
6 0

Answer:

Step-by-step explanation:

You didn't post the question!!

Domain: [ - 6, 3 )

Range: [ - 4, - 9 ]

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Which algebraic properties was applied to the expression a + (b + c) = (a + b) + c?
andrew11 [14]

Answer:

the Associative Property

Step-by-step explanation:

Notice that the ORDER doesn't change -- a, b, c on both sides.  But the GROUPING changes.

Example:

3 + (4 + 5) = (3 + 4) + 5

3 + 9 = 7 + 5

12 = 12

Subtraction does NOT behave this way!

8 - (7 - 1) = 8 - 6 = 2

(8 - 7) - 1 = 1 - 1 = 0

It matters who you associate with!  :-)

5 0
3 years ago
convert the following logarithmic equation to the equivalent exponential equation. Use the caret (^) to enter exponents. y=ln x
GarryVolchara [31]
1. To solve this problem, you need to remember that an exponential function has the following form:
 
 f(x)=a^x
 
 "a" is the base and "x" is the exponent.
 
 2. It is important to know that the logarithmic functions  and the exponential functionsare inverse. Then, you have:
 
 <span>y=ln x
</span> e^y=e^(lnx)
<span> e^y=x
 
 3. Therefore, the answer is:
</span> 
 x=
<span>e^y</span>
5 0
4 years ago
Please solve these following problems and show the work please and thanks! Photo for reference!
Sergeeva-Olga [200]

Step by step ' d4 34 495 04 959 504 40 59 495 9 49 20 150 10 495 40 50 95 309 4589 54

6 0
3 years ago
Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discr
Bad White [126]

Answer:

A quadratic equation can be written as:

a*x^2 + b*x + c = 0.

where a, b and c are real numbers.

The solutions of this equation can be found by the equation:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

Where the determinant is D = b^2 - 4*a*c.

Now, if D>0

we have the square root of a positive number, which will be equal to a real number.

√D = R

then the solutions are:

x = \frac{-b +- R }{2*a}

Where each sign of R is a different solution for the equation.

If D< 0, we have the square root of a negative number, then we have a complex component:

√D = i*R

x = \frac{-b +- C*i }{2*a}

We have two complex solutions.

If D = 0

√0 = 0

then:

x = \frac{-b +- 0}{2*a} = \frac{-b}{2a}

We have only one real solution (or two equal solutions, depending on how you see it)

3 0
3 years ago
Step by step <br> answer <br> please
Airida [17]

Answer:

C. 15

Step-by-step explanation:

<h2><em><u>∠ADC=∠ABC/2=30/2=15</u></em></h2>
7 0
3 years ago
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