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Anon25 [30]
2 years ago
11

A triangle has side lengths of 5 inches, 12 inches, and 15 inches. Every dimension is multiplied by 1/5 to form a new triangle.

How is the ratio of the perimeters related to the ratio of corresponding sides?
Mathematics
1 answer:
Butoxors [25]2 years ago
3 0

a = 5 in

b = 12 in

c = 15 in

the lengths of the sides a, b and c

the perimeter is P = a + b + c = 5 + 12 + 15 = 32 in

let the dimensions of the new triangle be

a1 = (1/5)*5 in

b1 = (1/5)*12 in

c1 = (1/5)*15 in

the perimeter is P1 = a1 + b1 + c1 = (1/5)5 + (1/5)12 + (1/5)15 = (1/5)(5 + 12 + 15) = (1/5)P1

P1 = (1/5)P

P1/P = 1/5 = a/a1 = b/b1 = c/c1

the ratio of the perimeters is equal to the ratio of the corresponding sides.

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5m-8=2m+2 <br><br> what does m equal
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Answer:

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<em />

Step-by-step explanation:

5m - 8 = 2m + 2

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 3m - 8 = 2    

      + 8 + 8              <em>(get m entirely on 1 side)</em>

3m = 10

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<em />

<em />

hope this helps!! have a lovely day :)

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1 year ago
Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

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so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

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