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V125BC [204]
3 years ago
7

PLEASE HELP ME ASAP, PLEASE I NEED HELP

Mathematics
1 answer:
kow [346]3 years ago
8 0

Answer:

Third one is correct

Step-by-step explanation:

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Grant spent 2.50,$4.00,$4.25, and $3.25?On breakfast one week.The next week he spent $6 more in total for the 4 breakfasts than
masha68 [24]

We have been given that Grant spent $2.50, $4.00, $4.25, and $3.25 on breakfast in one week. The next week he spent $6 more in total for the 4 breakfasts than the week before. We are asked to find increase in the mean of second week.  

Since Grant spent $6 more than last week, we will divide 6 by 4 to get how much mean of second week breakfast expenditures increased with respect to first week expenditures.

\text{Mean increase in second week breakfast expenditure}=\frac{6}{4}

\text{Mean increase in second week breakfast expenditure}=1.5

Therefore, mean of second week breakfast expenditure will be $1.5 more than first week.

3 0
3 years ago
∫c<br> x sin y ds, C is the line segment from (0, 1) to (3, 5)
Mnenie [13.5K]
Parameterize the line segment C by

\mathbf r(t)=\langle x(t),y(t)\rangle=(1-t)\langle0,1\rangle+t\langle3,5\rangle=\langle3t,1+4t\rangle

where t\in[0,1]. Then

\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt
\mathrm ds=\|\langle3,4\rangle\|\,\mathrm dt
\mathrm ds=5\,\mathrm dt

So the integral is

\displaystyle\int_Cx\sin y\,\mathrm ds=5\int_0^1x(t)\sin y(t)\,\mathrm dt
=\displaystyle5\int_0^13t\sin(1+4t)\,\mathrm dt
=\dfrac{15}{16}(\sin5-\sin-4\cos5)\approx-2.7516
3 0
3 years ago
How to solve this task???
Radda [10]
So (m+n)/2= 7

To find m+n, multiply both sides by 2
m+n=14

Final answer: C
3 0
3 years ago
Read 2 more answers
Without plotting these numbers, decide which inequalities are true. Check all that apply.
Alexeev081 [22]

Answer:

1,2,3,5

Step-by-step explanation:

<em>no cap fam </em>

8 0
3 years ago
Read 2 more answers
a) Suppose f"(z) exists on an interval I and f(s) has a zero at three distinct points a &lt; b&lt; c on I. Show there is a pbint
Fynjy0 [20]

Answer:

We can find a root in the average point (a+b+c)/3

Step-By-Step Explanation:

We can use the Rolle Theorem. Since f is two times deribable on both [a,b] and [b,c], then there exists points x in (a,b) and y in (b,c) such that f'(x) = f'(y) = 0. Now, again by using Rolle Theorem, since f' is derivable in [x,y] (because it is a closed interval in I), then there exists s in [x,y] such that f''(s) = 0. This proves a.

The cube (x-a)(x-b)(x-c) has 3 roots, a, b and c and it is two times derivable because it is a polynomial. Hence we can use (a) to ensure that there is a root on I. Nevertheless, we can try to find the root manually:

f'(x) = (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)

f''(x) = (x-c)+(x-b)+(x-c)+(x-a)+(x-b)+(x-a) = 2(x-a)+2(x-b)+2(x-c) = 2( (x-a) + (x-b) + (x-c) ) = 6x - 2 (a+b+c).

We want x such that

6x - 2(a+b+c) = 0, or, equivalently,

3x - (a+b+c) = 0

Hence

x = (a+b+c)/3

Is a root of f'' in I.

3 0
3 years ago
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