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Anika [276]
3 years ago
8

Addison earn $25 mowing her neighbor's lawn then she loaned her friend $18 and got 50 from her grandmother for her birthday she

now has $86 how much money did Addison have to begin with
Mathematics
2 answers:
Flura [38]3 years ago
6 0
She had $29. You just do the equation backwards. 86-57=29. 
To check if your answer is correct;
29+25=54
54-18=36
36+50=86
yarga [219]3 years ago
4 0

Solution: The total money did addison have to begin with is $29.

Explanation:

Let the addison begin with money $x.

It is given that the addison earn $25 from his neighbor and $50 from her grandmother. So total money addison has \$x+\$25+\$50=\$x+\$75

She loaned her friend %18.

Now she has \$x+\$75-\$18=\$x+\$57

It is given that she had $86.

\$x+\$57=\$86\\\$x=\$86-\$57\\\$x=\$29

Therefore, the total money did addison have to begin with is $29.

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Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki
Digiron [165]

Answer:

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for f(x) = 3 - \dfrac{1}{2}x from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x

where:

\Delta x = \dfrac{b-a}{a}

a = 2

b =14

n = 6

∴

\Delta x = \dfrac{14-2}{6}

\Delta x = \dfrac{12}{6}

\Delta x =2

Hence;

x_0 = 2 \\ \\  x_1 = 2+2 =4\\ \\  x_2 = 2 + 2(2) \\ \\  x_i = 2 + 2i

Here, we are  using left end-points, then:

x_i-1 = 2+ 2(i-1)

Replacing it into Riemann equation;

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix}  2+2 (i-1)  \end {pmatrix} \end {pmatrix}2

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2i-2)

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -   \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - 2  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} i

Estimating the integrals, we have :

= 6n - 2 ( \dfrac{n(n-1)}{2})

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

5 0
3 years ago
Solve for x 3x-1=9x+2
ANEK [815]
3x - 1 = 9x + 2

3x (-3x) - 1 (-2) = 9x (-3x) + 2 (-2)

-3 = 6x

-3/6 = 6x/6

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3 years ago
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2 years ago
Caroline bakes brownies to give to her friends. She starts with 48 brownies and gives a brownie to 6 friends each hour until she
Dmitrij [34]

Answer: The  below figure shows the graph of the given situation.

Step-by-step explanation:

Since, at first Caroline has brownies= 48

And, according to the question, it is decreasing by 6 per hour.

Then, the function of the given situation can be written as,

f(x)=48 - 6 x,     ------ (1)

where,  x is the number of hours and f(x) is the number of brownies Caroline has left in x hour.

Thus by substitution x= 0,1,2,3,4,5,6,7,8 in equation (1),

We get, f(x)=48, 42, 36, 30, 24, 18, 12, 6, 0

So, the function has points, (0, 48), (1, 42), (2, 36), (3, 30), (4, 24), (5,18 ),(6,12 ) (7, 6) (8, 0).

With help of the above points we can draw the graph of the given situation.

Note: since at x=8, f(x)=0 therefore after 8 hours she has no brownies left.


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Answer:

n + q = 28 0.05n + 0.25q = 4

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