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Alexxandr [17]
2 years ago
8

The extremes in the proportion 3/4=15/20 are

Mathematics
2 answers:
satela [25.4K]2 years ago
6 0
A proportion is an equation where two ratios are equal and a proportion has given names and these are the extremes and the means. The extreme and the mean allows you to do cross multiplying. In the given proportion above 3/4 is the same with 3 : 4 and 15/20 is 15 : 20. The extremes are 3 and 20, whereas the means are 4 and 15. Therefore the correct answer would be the second option.
Sonbull [250]2 years ago
4 0

Answer:

The answer is 3/20

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59.5 cm2 is the answer.
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Locate the points of discontinuity in the piecewise function shown below.
algol13

Answer:

Step-by-step explanation:

The given piecewise function i

From the given function it is clear that function is divided at x=-1 and x=2. It means we check the discontinuity at x=-1 and x=2.

For x=-1,

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Since LHL ≠ f(-1), therefore the given function is discontinuous at x=-1.

For x=2,

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Since LHL ≠ f(2), therefore the given function is discontinuous at x=2.

Therefore, the correct option is A.

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Consider the following system of equations. StartLayout Enlarged left-brace 1st row negative 10 x squared minus 10 y squared = n
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These 2 equations has no solution and the equations are independent of each other.

<u>Step-by-step explanation:</u>

-10x² -10y² = -300 ----a

5x² + 5y² = 150 ---- b

While trying to solve this,

We can multiply the eq. b by 2 so we will get eq. c and then add to eq. a we will get 0 as the solution.

10x² + 10y² = 300 ----c

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Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
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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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just add the 2 numbers together,

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3 years ago
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