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konstantin123 [22]
3 years ago
14

Use linear approximation to approximate √25.3 as follows.

Mathematics
1 answer:
Sophie [7]3 years ago
4 0

The idea is to use the tangent line to f(x)=\sqrt x at x=25 in order to approximate f(25.3)=\sqrt{25.3}.

We have

f(x)=\sqrt x\implies f(25)=\sqrt{25}=5

f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}

so the linear approximation to f(x) is

L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92

Hence m=\frac1{10} and b=\frac92.

Then

f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}

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James gave Robert half a candy bar. Robert gave his sister half of what he had. What fraction of the whole candy bar did Robert
Evgesh-ka [11]

Answer:

1/4

25%

Step-by-step explanation:

Robert has 1/2 of the candy bar

If he gave his sister half, she has 1/2 x 1/2 = 1/4 of the candy bar

To convert 1/4 to percentage, multiply by 100

1/4 x 100 = 25%

7 0
3 years ago
Prove for any positive integer n, n^3 +11n is a multiple of 6
suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0
3 years ago
Read 2 more answers
What is the order of operations to solve this problem?<br> 10 x 5 + 2 / 2 = 60
Slav-nsk [51]

Simply remember PEMDAS:

P = Parentheses

E = Exponents

M = Multiplication

D = Division

A = Addition

S = Subtraction

You perform these actions from left to right.

However in this case it is not possible for it to equal 60.

8 0
3 years ago
Read 2 more answers
Tina is saving to buy a notebook computer. She has two options. The first option is to put $500 away initially and save $30 ever
Jlenok [28]
First option:
y=money saved
x=number of months

y=30x+500

Second option:

y=50x+200

We have this system of equations:

y=30x+500
y=50x+200

We can solve this system of equations by equalization method
30x+500=50x+200
30x-50x=200-500
-20x=-300
x=-300/-20
x=15

so;
y=30x+500
y=30(15)+500
y=450+500
y=950

Answer; after 15 months, she would save the same amount using either option, the amount saved in either option will be $950. 
3 0
3 years ago
Determine
mihalych1998 [28]

Answer:

=4.+2.

Step-by-step explanation:

<u>Linear Combination Of Vectors </u>

One vector \vec b is a linear combination of \vec a_1 and \vec a_2 if there are two scalars x_1, x_2 such as

\vec b=x_1\vec a_1+x_2\vec a_2

In our case, all the vectors are given in R^3 but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.

We have

\vec a_1=,\ \vec a2=,\ \vec b=

We set the equation

=x_1.+x_2.

Multiplying both scalars by the vectors

=+

Equating each coordinate, we get

4x_1-4x_2=8

5x_1+3x_2=26

-4x_1+3x_2=-10

Adding the first and the third equations:

-x_2=-2

x_2=2

Replacing in the first equation

4x_1-4(2)=8

4x_1=8+8

x_1=4

We must test if those values make the second equation become an identity

5(4)+3(2)=20+6=26

The second equation complies with the values of x_1 and x_2, so the solution is

=4.+2.

8 0
3 years ago
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