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Katyanochek1 [597]
2 years ago
7

What is the fifteenth term of the sequence

Mathematics
1 answer:
eduard2 years ago
6 0

Answer:

81920.

Step-by-step explanation:

1) the ration of the given sequence is -2=-10/5=20/(-10)...

2) the 15th term can be calculated according to the formula:

a₁₅=a₁*(ratio)¹⁴, then

a₁₅=5*(-2)¹⁴=-5*1024*16=81920.

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The product of a number m and 6 is no less than 30.
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Answer:

x= anything less than 5

Step-by-step explanation:

We know that 5*6 is 30, so x would have to be anything less than five.

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Ravi drove 871 miles in 13 hours.
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Y= 4x-9 <br> and y=x-3 <br><br><br> Also using transitive property
Andreyy89
Y = 4x - 9 and y = x-3 so the two right sides (4x-9 and x-3) must be equal

4x-9 = x-3

Let's solve for x

4x-9 = x-3
4x-9+9 = x-3+9 ... add 9 to both sides
4x = x+6
4x-x = x+6-x ... subtract x from both sides
3x = 6
3x/3 = 6/3 ... divide both sides by 3
x = 2

If x = 2, then y is...
y = 4*x - 9 = 4*2-9 = 8-9 = -1
which can also be found by 
y = x-3 = 2-3 = -1

So overall x = 2 and y = -1

We have the ordered pair (x,y) = (2,-1)
4 0
3 years ago
PLEASE HELP ME:<br><br> Find the perimeter of the figure. Round to the nearest tenth.
Ksenya-84 [330]

Answer:

perimeter = 20.9 units

Step-by-step explanation:

perimeter

perimeter = distance around two dimensional shape

= addition of all sides lengths

<h2>perimeter of the figure</h2><h2>= AB+BC+CD+AD</h2>

distance formula:

d =  \sqrt{(  x_{2} -  x_{1})  {}^{2}   + ( y_{2} -  y_{1}) {}^{2}  }

<h3>1) distance of AB</h3>

A(-3,0) B(2,4)

x1 = -3 x2 = 2

y1 = 0 y2 = 4

(substitute the values into the distance formula)

ab =  \sqrt{(2 - ( - 3)) {}^{2}  + (4 - 0) {}^{2} }

ab =  \sqrt{5 {}^{2} + 4 {}^{2}  }

ab =  \sqrt{41}AB = 6.4 units

<h3>2) distance of BC</h3>

B(2,4) C(3,1)

x1 = 2 x2 = 3

y1 = 4 y2 = 1

bc =  \sqrt{(3 - 2) {}^{2}  + (1 - 4) {}^{2} }

bc = \sqrt{1 {}^{2} + ( - 3) {}^{2}  }

bc =  \sqrt{10}

BC = 3.2 units

<h3>3) distance of CD</h3>

C(3,1) D(-4,-3)

x1 = 3 x2 = -4

y1 = 1 y2 = -3

cd =  \sqrt{( - 4 - 3) {}^{2}  + ( - 3 - 1)) {}^{2} }

cd =  \sqrt{( -   7) {}^{2}  + ( - 4 ){}^{2} }

cd =   \sqrt{65}

CD = 8.1 units

<h3>4) distance of AD</h3>

A(-3,0) D(-4,-3)

x1 = -3 x2 = -4

y1 = 0 y2 = -3

ad =  \sqrt{(  - 4 - ( - 3)) {}^{2}  +  ( - 3 - 0) {}^{2} }

ad =  \sqrt{( - 1) {}^{2} + ( - 3) {}^{2}  }

ad =  \sqrt{10}

AD = 3.2 units

<h2>perimeter of figure</h2>

= AB+BC+CD+AD

= 6.4 + 3.2 + 8.1 + 3.2

= 20.9 units

8 0
3 years ago
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