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beks73 [17]
3 years ago
10

Decide if the following is a true or false statement - explain how you know (Explain in words please).

Mathematics
2 answers:
Westkost [7]3 years ago
7 0

Answer:

it's negative and positive and negative + positive is -ive or +ive but in this case they are in equal distance from the number line which makes the answer zero

Yakvenalex [24]3 years ago
5 0
It is negative and positive and the answer is zero
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Find the circumference of a circle with a radius of 10 m. C = 2πr, where d=diameter and C = circumference. Use pi = 3.14.
Tju [1.3M]

Answer:

3.14×20

20=diameter

=62.8cm

4 0
3 years ago
Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0
3 years ago
Evaluate the following expression using the values given: (1 point)
jolli1 [7]
3(-2) - 1 - 3(-2) = -1
5 0
3 years ago
The graph of an exponential function has a y-intercept of 8 and contains the point (3,64). Find the exponential function that de
Brums [2.3K]

Given:

Y-intercept of exponential function is 8.

It contains the point (3,64).

To find:

The exponential function that describes the graph.

Solution:

The general form of an exponential function is

y=ab^x      ...(i)

where, a is initial value or y-intercept and b is growth factor.

Since, y-intercept is 8, therefore, a=8.

Put a=8 in (i).

y=8b^x         ...(ii)

It contains the point (3,64). Put x=3 and y=64.

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Divide both sides by 8.

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8=b^3

2^3=b^3

On comparing both sides, we get

b=2

Put b=2 in (ii).

y=8(2)^x

The functions form of this equation is

f(x)=8(2)^x

Therefore, the required function is f(x)=8(2)^x.

6 0
3 years ago
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4,872 square miles larger

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