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BigorU [14]
3 years ago
6

Transform the equation if necessary, and then solve it to find the value of x that makes the equation true.

Mathematics
2 answers:
Mrrafil [7]3 years ago
4 0

Answer:

X = 1.030714

Step-by-step explanation:

icang [17]3 years ago
3 0

Answer:

x = 10/7 = 1.429

Step-by-step explanation:

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Write an equation for the line that is parallel to the given line and that passes through the given point. y = –6x + 2; (–1, 2)
kap26 [50]
Parallel=same slope
Y=mx+b
Sub in the slope and point to solve for b
2=-6(-1)+b
2=6+b
2-6=b
-4=b
The equation is y=-6x-4; the answer is B
Hope this helps!
6 0
3 years ago
This lesson is almost ⅔ over. Express this number as a decimal.
MissTica

Answer:

0.66666667

Step-by-step explanation:

7 0
3 years ago
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Nathan packs 30 boxes in 2 hours.how many boxes can he pack in an 10 hour shift
Advocard [28]

Question :

nathan packs 30 boxes in 2 hours.how many boxes can he pack in an 10 hour shift?

30 boxes in 2 hours = 15 box in 1 hour

so

10 * 15 = 150

150 boxes is your answer

6 0
3 years ago
Read 2 more answers
How to solve these 3 problems
Bingel [31]

Answer:

  1. a. decay; b. growth; c. decay; d. neither
  2. r = 4; a = 1; y = 1·4^x
  3. a. an = 3(5^(n-1)); b. f(x) = (3/5)(5^x); c. exponential growth; d. y-intercept: 3/5; first term: 3.

Step-by-step explanation:

There are two kinds of exponential problems here.

  1. exponential functions of the form f(x) = a·b^x
  2. exponential sequences of the explicit form an = a1·r^(n-1)

The second problem gives you a table that suggests the sequence form, but it asks for the exponential function form. The third problem does something similar.

__

<h3>1.</h3>

In an exponential function of the form f(x) = a·b^x, the function grows if b>1 and decays if b<1. Using this check, we can easily answer ...

  a. 0.4 < 1 . . . decay

  b. 1.3 > 1 . . . growth

  c. 1/2 < 1 . . . decay

  d. 1 = 1 . . . neither growth nor decay; the function is constant: j(x) = 1.

__

<h3>2.</h3>

The value of x is given starting at 1, so we can consider this a geometric sequence. The common ratio is r = 16/4 = 4. The first term is a1 = 4, so the explicit formula for the sequence is ...

  an = 4·4^(n-1)

When this is expanded to get rid of the constant in the exponent, we have ...

  an = 4·(4^n)·(4^-1) = 1·4^n

We recognize this form as matching the functional form f(x) = a·r^x. The multiplier of the exponential factor is a=1. In summary, ...

  r = 4; a = 1; f(x) = 1·4^x

__

<h3>3.</h3>

The first term of this geometric sequence is a1 = 3. The common ratio is r = 15/3 = 5. Using the explicit formula, we have ...

 a. explicit form: an = 3·5^(n-1)

Using the method of question 2 to write the functional form, we find ...

  an = 3(5^n)(5^-1) = (3/5)(5^n)

  b. functional form: f(x) = (3/5)(5^x)

  c. function family: exponential growth functions

 d. y-intercept: (3/5) . . . . read this from the f(x) form

     1st term: the first term listed in the given sequence is 3

_____

<em>Additional comment</em>

The "y-intercept" of a sequence is irrelevant (undefined), as the sequence term numbering starts with 1, not 0. The domain of the explicit formula is <em>natural numbers</em>, which does not include 0.

Similarly, the "first term" of a function f(x) needs further definition. Here, we've answered the question by saying the first term is f(1). There is no conventional definition of a "first term" for a continuous function.

3 0
2 years ago
I need help really bad please
yuradex [85]

Answer:

9

Step-by-step explanation:

72 = 2^3 \cdot \boxed{3^2}

27 = 3 \cdot \boxed{3^2}

3 0
2 years ago
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