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Maslowich
3 years ago
10

calculate the distance between the points!

Mathematics
2 answers:
OLEGan [10]3 years ago
4 0

Answer:

The answer is the distance is 9. I am hopefully correct

weeeeeb [17]3 years ago
4 0

Distance formula: d = √(x2-x1)^2+(y2-y1)^2

d = √(8-1)^2+(-9-(-1)^2)

d = √7^2+(-8)^2

d = √49+64

d = √113

d ≈ 10.63

Best of Luck!

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The sum of the first four terms of an arithmetic sequence is $10$. If the fifth term is $5$, what is the sixth term?
Nookie1986 [14]

The sixth term of an arithmetic sequence is 6

<h3>How to find arithmetic sequence?</h3>

The sum of the first four terms of an arithmetic sequence is 10.

The fifth term is 5.

Therefore,

sum of term = n / 2(2a + (n - 1)d)

where

  • a = first term
  • d = common difference
  • n = number of terms

Therefore,

n = 4

10 = 4 / 2 (2a + 3d)

10 = 2(2a + 3d)

10 = 4a + 6d

4a + 6d = 10

a + 4d = 5

4a + 6d = 10

4a + 16d = 20

10d = 10

d = 1

a + 4(1) = 5

a = 1

Therefore,

6th term = a + 5d

6th term = 1 + 5(1)

6th term = 6

learn more on sequence here: brainly.com/question/24128922

#SPJ1

7 0
1 year ago
Does anyone know how to Graph f(x)=51(2)^x
3241004551 [841]
This is an exponential function.   

If x = 0, 2^x = 2^0 = 1.  The beginning value of 2^x is 1 and the beginning value of 51*2^x is 51.

Make a table and graph the points:

x        y=51*2^x                                point (x,y)
--       ---------------                            ---------------
0             51                                       (0,51)
2            51*2^2 = 51(4) = 204           (2,204)             and so on.

The graph shows up in both Quadrants I and II.  Its y-intercept is (0,51).  Its slope is always positive.

5 0
3 years ago
Read 2 more answers
Can someone help me with this question please
DENIUS [597]
37 im pretty sure sorry if not correct.
6 0
3 years ago
Hlol can u plz solve b no plz plz.....I will mark as brainliest​
zimovet [89]

Answer:

Step-by-step explanation:

Tan \ B =\frac{4}{3} = \frac{opposite \ side \ of B}{adjacent \ side \ of B}

BC = 15 cm

3x = 15

x = 15/3 = 5

AB =4x = 4*5 = 20 cm

Pythagorean theorem,

AC² = AB² + BC²

       = 20²  + 15 ²

       = 400 + 225

       = 625

AC = √625 = 25 cm

AC = 25 cm

Sin \ A = \frac{opposite  \ side \ of \ angle A }{hypotenuse}\\\\  = \frac{BC}{AB}=\frac{15}{25}=\frac{3}{5}

6 0
3 years ago
The segments shown below could form a triangle.
erastovalidia [21]

sorry i don,t unerstan what u wn,t

8 0
3 years ago
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