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3 years ago

A triangle has side lengths of 3 feet and 9 feet.which is the greatest possible perimeter of the triangle

Mathematics

1 answer:

OLga [1]3 years ago

6 0

21 because you have to have at least 1 side as 3m so the other two should be 9m to have the greatest possible length

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Which of the following

Ronch [10]

Answer:

Option A

Step-by-step explanation:

Given equation is

=> 3y = 6x + 3

In slope-intercept form, it becomes

=> 3y = 3(2x+1)

=> y = 2x+1

So, Slope = m = 2

Parallel lines have equal slope, So any line parallel to the above line would have its slope equal to 2

=> Line parallel to 3y = 6x + 3 is y = 2x + 10

3 0

4 years ago

HELP PLEASE 21 POINTS PLEASE HELP ME!!!! <3

Anna007 [38]

n 1 2 3 4 5 6

f(n) 1033 932 831 730 629 528

First term (a₁): 1033 Common difference (d): -101

Explicit rule: $a_{n} = 1134 - 101n$ Recursive rule: $a_{n} = a_{n-1} - 101$

$a_{n} = a_{1} + d(n - 1)$

$a_{n} = 1033 - 101(n - 1)$

$a_{n} = 1033 - 101n + 101$

$a_{n} = 1134 - 101n$

***********************************************************************************

n 1 2 3 4 5 6

f(n) -39 -29 -19 -9 9 19

First term (a₁): -39 Common difference (d): +10

Explicit rule: $a_{n} = -49 + 10n$ Recursive rule: $a_{n} = a_{n-1} + 10$

$a_{n} = a_{1} + d(n - 1)$

$a_{n} = -39 + 10(n - 1)$

$a_{n} = -39 + 10n - 10$

$a_{n} = -49 + 10n$

***********************************************************************************

n 1 2 3 4 5 6

f(n) 3.75 2.5 1.25 0 -1.25 -2.5

First term (a₁): 3.75 Common difference (d): -1.25

Explicit rule: $a_{n} = 5 - 1.25n$ Recursive rule: $a_{n} = a_{n-1} - 1.25$

$a_{n} = a_{1} + d(n - 1)$

$a_{n} = 3.75 - 1.25(n - 1)$

$a_{n} = 3.75 - 1.25n + 1.25$

$a_{n} = 5 - 1.25n$

6 0

4 years ago

What is the equation, of the perpendicular to y= 1/8x + 2, , passing through the point (1,-5)?

BARSIC [14]

Answer:

yes passing is the best thing to do so you can get a job

Step-by-step explanation:

7 0

3 years ago

Sixteen-twentyfive as a decimals

olga_2 [115]

16.25 or .1625 is the awsner

8 0

3 years ago

Find the dimensions of the rectangular box with largest volume if the total surface area is given as 64cm^2

motikmotik

Let x =lenght, y = width, and z =height
The volume of the box is equal to V = xyz 
Subject to the surface area 
S = 2xy + 2xz + 2yz = 64 
= 2(xy + xz + yz) 
= 2[xy + x(64/xy) + y(64/xy)] 
S(x,y)= 2(xy + 64/y + 64/x) 
Then 
Mx(x, y) = y = 64/x^2 
My(x, y) = x = 64/y^2 
y^2 = 64/x 
(64/x^2)^2 = 64 
4096/x^4 = 64/x 
x^3 = 4096/64 
x^3 = 64 
x = 4 
y = 64/x^2 
y = 4 
z= 64/yx 
z= 64/16 
z = 4 

Therefor the dimensions are cube 4.

6 0

4 years ago