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kumpel [21]
3 years ago
14

If Emma has all a's and one d what is her gpa? she has 6 classes in total

Mathematics
1 answer:
olga_2 [115]3 years ago
3 0

Answer:

3.5 GPA

Step-by-step explanation:

I used a GPA calculator.

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PLZ HELP IM IN DESPERATE NEED OF AN ANSWER!!!
Irina18 [472]
I believe 4(x - 3)=32 could help I haven’t done an equation from the unit in months so hopefully that looks familiar to you.
5 0
3 years ago
A- 13<br> B- 26 <br> C- 52<br> D- there is not enough information
NISA [10]
The answer is A, 13. Since we know line DB is a median (the middle), it tells us that line BC must equal line AB. Simply just do 26/2.
6 0
3 years ago
if buyer purchases 25% of a lot in the spring and then purchases 50% more in the fall what percentage of the lot is still availa
brilliants [131]

Answer:

25%

Step-by-step explanation:

Total purchase = 25%+50%

Total purchase = 75%

Percentage left = 100% - 75%

Percentage left = 25%

8 0
3 years ago
The piecewise function h(x) is shown on the graph.
Artemon [7]

Step-by-step explanation/Answer:

<u><em>The value for h(3) is 1.</em></u>

<u><em>Coordinate (3,1)</em></u>

The value on h(3) is located in the last part of the piecewise function, which starts at (1,3) until (4,0). To know the exact image of h(3), we can find the function that belongs to that piece of line, and then calculated the asked value.

So, to calculate the equation, we first have to find the slope with its definition, and then we'll use the point-slope formula to find the equation:

<u><em>Therefore, the coordinates is (3,1), that is, the value for h(3) is 1.</em></u>

3 0
3 years ago
Read 2 more answers
What is the nth term?
kumpel [21]

Let a_k denote the <em>k</em>th term of the sequence. Then

a_k=a_1+d(k-1)

where <em>d</em> is the common difference between consecutive terms in the sequence and <em>a</em>₁ is the first term.

The sum of the first <em>n</em> terms is

S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n

From the formula for a_k, we get

S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)

S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k

S_n=na_1+\dfrac{d(n-1)n}2

S_n=\dfrac n2(2a_1-d+dn)

So we have d=-5, and 2a_1-d=16 so that a_1=\frac{11}2.

Then the <em>n</em>th term in the sequence is

a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}

7 0
3 years ago
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