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Anni [7]
3 years ago
9

Lisa and her husband are each starting a saving plan. Lisa will initially set aside

Mathematics
1 answer:
Lelechka [254]3 years ago
5 0

Please work on your formatting.


If this lady starts out with $125 and then saves an additional $50.85 per week after that, the formula for (her) savings is


S(x) = $125 + ($50.85 / week) x, where x is the # of weeks.


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The difference between twice a number and 2
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2 is a number and "twice a number" means multiply by 2.

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Determine whether theses lines are parallel, perpendicular or neither
belka [17]

Answer:

Lines 2 and 3 are parallel with slope (-2/3) and different y intercepts, and both are perpendicular to line 1              (-1) / (3/2) = (-2/3)

Step-by-step explanation:

Parallel lines have the same slope

Perpendicular lines have negative reciprocal slopes

line 1: 6x - 4y = 2

line 1: 4y =  6x - 2

line 1:   y = (3/2)x - 0.5

line 2:  y = (-2/3)x - 6

line 3:3y = -2x + 4

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8 0
3 years ago
What is the domain and derivative of f(x)=ln(2xsqrt(2+x))?
harkovskaia [24]
Domain is the numbers yo can use
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0


deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)


so

dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))



domain is all real numbers greater than 0
derivitive is \frac{3x+4}{2x(x+2)}
7 0
3 years ago
For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10
professor190 [17]

The true statements are:

  • Tyson's expression is not equivalent to the original expression
  • The equivalent expression is:\frac{1}{2}g- \frac 1{5}

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}

Collect like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10}  -\frac{1}{10}

Evaluate the like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}

Tyler's equivalent expression is given as:

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}

Collect like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}

Evaluate the like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

3 0
2 years ago
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