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torisob [31]
2 years ago
5

SOMEBODY PLEASE HELP ME

Mathematics
2 answers:
Ber [7]2 years ago
5 0

by Pythagoras theorem,

hypotenuse² = base² + adjacent²

x² = 7² + 11²

x² = 49 + 121

x² = 170

x = √170

x = 13.03

x = 13(round fig)

hope it helps...!!!

In-s [12.5K]2 years ago
4 0
A^2 + B^2 = C^2

7^2 + 11^2 = C^2

49 + 121 = C^2

170 = C^2

Root 170 = line bc
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Ms. Costco was offered the best price. It would be $15,900.57 after tax

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The sum of two numbers is 100. Five times the smaller number is eight more than the larger
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Answer:The larger of two numbers is 8 more than 4 times the smaller if the larger number is increased by four times the smaller the result is 40. From this statement, we can calculate the values of the numbers where you get the value of the larger number to be 24 and that of the smaller number to be 4.

Step-by-step explanation:

4 0
2 years ago
C+11=24 <br> please help!!
Ede4ka [16]

Answer:

13

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Parisian sewer #1 contains 9 fewer gigantic rats—rats as big as cats, some might say—than parisian sewer #2. parisian sewer #2 c
Leya [2.2K]

There are different kinds of math problem. There will be 11 rats in sewer #1.

<h3>What are word problem?</h3>

The term  word problems is known to be problems that are associated with a story, math, etc. They are known to often vary in terms of technicality.

Lets take

sewer #1 = a

sewer #2 = b

sewer #3 = c

Note that   A=B-9

So then you would have:

A=B-9

B=C- 5

A+B+C=56

Then you have to do a substitution so as to find C:

(B- 9) + (C-5) + C = 56

{ (C- 5)-9} + (C-5) + C = 56

3C - 19 = 56

3C = 75

B = C- 5

B = 25 - 5

Therefore, B = 20

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= 25 - 9

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Learn more about  Word Problems from

brainly.com/question/21405634

8 0
2 years ago
Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

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c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
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