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damaskus [11]
4 years ago
5

What is the answer? Please help

Mathematics
1 answer:
pav-90 [236]4 years ago
7 0

Answer:

Step-by-step explanation:

g(7) = 7^2 - 3(7) = 49 - 21 = 28

f(28) = 4(28) + 4 = 112 + 4 = 116

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Which statement is true about the slope of the graphed line?
Galina-37 [17]
It’s B ) THE SLOPE IS POSITIVE
6 0
4 years ago
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Help please and thank you no bots please
I am Lyosha [343]

Answer:

Step-by-step explanation:

midline: any point along the line y = -5

- (-2pi, -5)

- (0, -5)

- (2pi, -5)

maximum: (pi, -2)

minimum: (-pi, -8)

8 0
3 years ago
A cargo helicopter delivers only 100100100-pound packages and 120120120-pound packages. For each delivery trip, the helicopter m
Romashka [77]

Answer:

5

Step-by-step explanation:

Given that:

Weight of package delivered = 100 - 120 pounds

Number of packages carried per trip is atleast 10 = ≥ 10

Total weight of packages must be at most 1,100 ≤1,100

The maximum number of 120 pound packages the helicopter can carry:

Maximum capacity = 1100 pounds

Number of packages ≥ 10

Let number of 110 packages carried = x

Number of 120 packages carried = y

x + y ≥ 10 - - (1)

100x + 120y ≤ 1100 - - - (2)

From (1): x ≥ 10 - y

100(10 - y) + 120y ≤ 1100

1000 - 100y + 120y ≤ 1100

1000 + 20y ≤ 1100

20y ≤ 1100 - 1000

20y ≤ 100

y = 5

y = number of 120 pound packages carried

y = 5

8 0
3 years ago
Read 2 more answers
Find the simple interest earned by each account. $800 principal at 4% interest for 5 years
Kipish [7]

Answer:

$160

Step-by-step explanation:

Simple interest=principal×time×rate/100

Principal=$800

Time=5years

Rate=4%

S.I=800×5×4/100

S.I=8×4×5

S.I=160

So the simple interest is $160

6 0
3 years ago
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Find the sum Sn below:
Hoochie [10]

We can write <em>S</em> as

\displaystyle S = \sum_{k=0}^{n-1} (n-k)3^k

and expand it as

\displaystyle S = n \sum_{k=0}^{n-1} 3^k - \sum_{k=0}^{n-1} k\cdot3^k

The first sum is geometric, nothing tricky:

\displaystyle\sum_{k=0}^{n-1} 3^k = 1 + 3 + 3^2 + \cdots + 3^{n-1} \\\\ \implies 3\sum_{k=0}^{n-1} 3^k = 3 + 3^2 + 3^3 + \cdots + 3^n \\\\ \implies -2\sum_{k=0}^{n-1} 3^k = 1 - 3^n \\\\ \implies \sum_{k=0}^{n-1} 3^k = \frac{3^n-1}2

For the second sum, you can use the same method employed in another question of yours (24494877) to find

\displaystyle \sum_{k=0}^{n-1} k\cdot 3^k = \frac{(2n-3)3^n+3}4

So this sum comes out to

\displaystyle S = n\cdot\frac{3^n-1}2 - \frac{(2n-3)3^n+3}4 \\\\ S = \frac{2n\cdot3^n-2n - (2n-3)3^n-3}4 \\\\ \boxed{S = \frac{3^{n+1}-2n-3}4}

7 0
3 years ago
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