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Alina [70]
3 years ago
15

What is the value of the expression 1 – r – a when r = 20 and a = 6?

Mathematics
2 answers:
Studentka2010 [4]3 years ago
6 0
B. -25, because 1 minus 20 is -19, then you subtract 6 from -19, then the answer is -25.
rodikova [14]3 years ago
4 0
1 - r - a      r=20    a =6
1 - 20 - 6 
  -19 - 6 =  -19 + (-6) 
     -25 would be the answer 


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What is the lcd of (3x)/(x + 3), 3/(x ^ 2 - 9)
Rudik [331]

Answer:

Simplify the denominator.

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x

(

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Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

(

x

+

3

)

(

x

−

3

)

,

(

x

−

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−

2

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The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number

1

is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of

1

,

1

is the result of multiplying all prime factors the greatest number of times they occur in either number.

1

The factor for

x

+

3

is

x

+

3

itself.

(

x

+

3

)

=

x

+

3

(

x

+

3

)

occurs

1

time.

The factor for

x

−

3

is

x

−

3

itself.

(

x

−

3

)

=

x

−

3

(

x

−

3

)

occurs

1

time.

The factor for

x

−

2

is

x

−

2

itself.

(

x

−

2

)

=

x

−

2

(

x

−

2

)

occurs

1

time.

The LCM of

x

+

3

,

x

−

3

,

x

−

3

,

x

−

2

is the result of multiplying all factors the greatest number of times they occur in either term.

(

x

+

3

)

(

x

−

3

)

(

x

−

2

)

Step-by-step explanation:

there does that help

6 0
3 years ago
If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)
Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

\log_dabc=\log_da+\log_db+\log_dc

Then using the change of base formula, we can derive the relationship

\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}

This immediately tells us that

\log_dc=\dfrac1{\log_cd}=\dfrac12

Notice that none of a,b,c,d can be equal to 1. This is because

\log_1x=y\implies1^{\log_1x}=1^y\implies x=1

for any choice of y. This means we can safely do the following without worrying about division by 0.

\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}

so that

\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23

Similarly,

\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}

so that

\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34

So we end up with

\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}

###

Another way to do this:

\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}

\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}

\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12

Then

abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}

So we have

\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}

4 0
3 years ago
Topic: Absolute Value Equations and Inequalities
jolli1 [7]

The values of  x are -22 and -2 and there are not extraneous solutions

<h3>How to solve the equation?</h3>

The equation is given as:

2|x + 7|= x - 8

Expand the absolute bracket

|2x + 14|= x - 8

Remove the absolute bracket

2x + 14 = x - 8 and 2x + 14 = -x + 8

Evaluate the like terms

x = -22 and 3x = -6

This gives

x = -22 and x = -2

Hence, the values of  x are -22 and -2 and there are not extraneous solutions

Read more about absolute value expressions at:

brainly.com/question/24368848

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4 0
2 years ago
At the movie theatre, child admission is $5.20 and adult admission is $8.50. On Sunday, three times as many adult tickets as chi
Gemiola [76]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Solve for x:<br><br>a) x=70<br>b) x=180<br>c) x=40<br>d) x=90​
Serga [27]

Answer:

a) x = 70

Step-by-step explanation:

70 (2) +20 +20

140 + 40

180

4 0
2 years ago
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