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den301095 [7]
2 years ago
3

The weight, in pounds, of a newborn baby tt months after birth can be modeled by the equation W=1.75t+6.W=1.75t+6. What is the s

lope of the equation and what is its interpretation in the context of the problem?
Mathematics
1 answer:
Alex Ar [27]2 years ago
7 0

Answer:

the slope is 1.75

Step-by-step explanation:

Interpretation: The weight of newborn is 6 lb at birth, and every month baby gained 1.75 lb. You replace x with month you want to know the weight.  For example you can calculate the weight after 4 month : w= 1.75(4)+6= 13 lb

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Let f (x) = –x2 + 3x + 1 and g(x) = 2x. Look at the composition shown. What is the error in the student’s work? f (g (x)) = –(2x
ioda

in term 3x, the correct is 3(2x)

5 0
2 years ago
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How is a calculation of net worth different from a day-to-day or month-to-month tallying of expenses?
alina1380 [7]
The difference in tallying of expenses from a day to day and month to month is the consistency. in a day to day basis it is consistent the time for a day is 24 hrs it will not change, so you will really know why the expenses goes up or down/ unlike for a month to month it is inconsistent,, some months have 30, 31 or 28 days in a month.
8 0
2 years ago
Write the first five terms of the geometric sequence in which a1=64 and the common ratio is 5/4
natita [175]

Remember that finding terms in a geometric sequence is done by multiplying the previous term by a common ratio r. For example, we can say:

a_2 = a_1 r

a_3 = a_2 r = (a_1 r)r = a_1 r^2


We have a_1 = 64. To find a_2, let's multiply this term by \frac{5}{4}:

a_2 = 64 \cdot \frac{5}{4} = 80


Now, let's use this to find all of our other terms:

a_3 = 80 \cdot \frac{5}{4} = 100

a_4 = 100 \cdot \frac{5}{4} = 125

a_5 = 125 \cdot \frac{5}{4} = \frac{625}{4}


Thus, our terms are 64, 80, 100, 125, and (625/4).

8 0
2 years ago
The balance on Taylor's credit card is $2000. It has an interest rate of 12.5%. She wants to compare the difference between payi
-Dominant- [34]

Answer: kindly check Explanation

Step-by-step explanation:

The balance on card = $2000

Interest rate = 12.5%

Therefore, monthly rate = 12.5% / 12 = 0.0104167

Paying $75 of monthly balance :

Interest amount on 2000 = 0.0104167 × 2000 = $20.83

Therefore,

Principal amount payed = 75 - 20.83 = 54.17

Balance = ($2000 + $20.83) - $75 = 1945.83

Interest on balance = 0.0104167 × 1945.83 = $20.2691

Paying $100 of monthly balance :

Interest amount on 2000 = 0.0104167 × 2000 = $20.83

Therefore,

Repayment = 100 - 20. 83 = $79.17

Balance = ($2000 + $20.83) - $100 = 1920.83

Interest on balance = 0.0104167 × 1920.83 = $20.0087

Interest saved by paying $100 instead of $75

$20.27 - $20.00 = $0.27 = 27 cent

3 0
2 years ago
Ella completed the following work to test the equivalence of two expressions. '
Romashka [77]

Answer:

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.

Step-by-step explanation:

<u>Equivalent algebraic expressions </u> are those expressions which on simplification give the same resulting<u> expression.</u>

Two algebraic <u>expressions</u> are said to be equivalent if their values obtained by substituting any values of the variables are same.

Two expressions 3f+2.6 and 2f+2.6 are not equivalent, because when f=1,

3f+2.6=3.1+2.6=3+2.6=5.6

2f +2.6=2.1+2.6=2+2.6=4.6

5.6≠   4.6

Method of substitution can only help her to decide the expresssions are not equivalent, but if she wants to prove the expressions are equivalent, she must prove it for all values of f.

3f+2.6=2f+2.6

3f=2f

3f-2f=0

f=0

This is true only when f=0.

Hence,

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.

this what i know

8 0
2 years ago
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