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Andreas93 [3]
3 years ago
5

Please help with this algebra :)

Mathematics
1 answer:
Dahasolnce [82]3 years ago
8 0
I am gonna go for D is the answer.
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This is an arithmetic series with common difference 6 

S13  = (13/2)[2*-5 + (13-1)*6)]  =  403
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Lim ln(tan x) as tends to pi/2 from the left
ioda
Let x=\arctan y. Then \tan x=y, and so as x\to\dfrac\pi2^-, you have y\to+\infty. The limit is then equivalent to

\displaystyle\lim_{x\to\frac\pi2^-}\ln(\tan x)=\lim_{y\to\infty}\ln y=\infty
3 0
3 years ago
write two equations that when graphed look like this *see image*. please help its due soon and i will give ya 75 points for answ
Zigmanuir [339]

The graph is showing the equation of circle x²+y²=4

<h3>What is a circle?</h3>

A circle is a two-dimensional geometry on the plane having a centre point and the circular line is drawn equidistant from the centre point.

The given graph in the question represents the equation of the circle cutting the points on the x-axis and y-axis at 4 which is the radius of the circle.

The equation will be as follows:-

x²+y²=4

Hence the graph is showing the equation of circle x²+y²=4

To know more about circles follow

brainly.com/question/24375372

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8 0
2 years ago
Can anyone please help!!!
quester [9]

Answer:

Question 9

(a) The distance, Jalaj walks in one day is 4.4 km

(b) The amount Jalaj raises after walking for 22 km at the end of the 5 days is $8

Question 10

(b) The difference between the largest and smallest areas is 2,400 m²

Step-by-step explanation:

Question 9

(a) The distance Jalaj walks in 5 days = 22 km

Whereby Jalaj walks equal distance every day, we have;

The distance, Jalaj walks in one day = 22 km/5 days = 4.4 km/day

The distance, Jalaj walks in one day = 4.4 km

(b) The amount he raises for every kilometer he walks = $1.60

The amount he raises after walking for 22 km at the end of the 5 days = 5 × $1.60  = $8

Question 10

(b) The given side length of the square = 120 meters to the nearest 10 meters

Therefore;

The maximum dimension for the side length of the square = 120 + 10/2 = 125

The largest possible area of the square, A_l = 125 m × 125 m = 15,625 m²

The minimum dimension for the side length of the square = 120 m - 10 m/2 = 115 m

The smallest possible area of the square, A_s = 115 m × 115 m = 13,225 m².

The difference between the largest and smallest areas, A_l - A_s = 15,625 - 13,225 = 2,400 m².

8 0
3 years ago
8. Graph the following linear equations.<br> a. 3x+5y=15
kenny6666 [7]

Answer:

slope: -3/5

y-intercept (0,3)

x    -      y

0          3

5          0

Step-by-step explanation:

I just did it

4 0
3 years ago
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