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Marina86 [1]
3 years ago
13

Is (0, 0) a solution to the equation y = 9x?

Mathematics
1 answer:
Levart [38]3 years ago
8 0

Answer:

Yes

Step-by-step explanation:

The equation y = 9x does intersect at the point (0,0) so therefore it is a solution.

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If Mrs. Ramsey divides 5 pounds of flour equally into three bowls, what is the weight of flour in one bowl? A) 3 5 pound Elimina
Mariulka [41]
Hi there!

The key in the question is "divide". Therefore, we divide 5 by 3

5/3=1.6 pounds

I don't think any of these are right.

3 0
3 years ago
Read 2 more answers
Samaria attached 165 inches of fringe material to a tablecloth if the tablecloth was in a shape of a regular pentagon how much f
Y_Kistochka [10]
35 in. because a pentagon has 5 sides and have to divide the total amount of yarn to the sides. 
4 0
3 years ago
hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a
nalin [4]

Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

6 0
3 years ago
A recipe for 5 batches of bagels uses 15 cups of flour. Ciro wants to know how many cups are needed for 1 batch. Johanna wants t
shutvik [7]

ciro needs 3 cups of flour johanna needs 20 x 15 cups

3 0
3 years ago
Read 2 more answers
Determine n between 0 and 19 such that (2311)(3912) ≡ n mod 20.
sleet_krkn [62]

You can write 2311 and 3912 in the form 20q+r:

2311=115\cdot20+11

3912=125\cdot20+12

Then

2311\cdot3912=(115\cdot20+11)(125\cdot20+12)

2311\cdot3912=115\cdot125\cdot20^2+(11\cdot125+12\cdot115)\cdot20+11\cdot12

Taken modulo 20, the terms containing powers of 20 vanish and you're left with

2311\cdot3912\equiv11\cdot12\equiv132\pmod{20}

We further have

132=6\cdot20+12

so we end up with

2311\cdot3912\equiv12\pmod{20}

and so n=12.

###

If instead you're trying to find 2311^{3912}\pmod{20}, you can apply Euler's theorem. We can show that \mathrm{gcd}(2311,20)=1 using the Euclidean algorithm. Then since \varphi(20)=8, and 8 divides 3912, we have

2311^{3912}\equiv2311^{489\cdot8}\equiv(2311^{489})^8\equiv1\pmod{20}

To show 2311 and 20 are coprime:

2311 = 115*20 + 11

20 = 1*11 + 9

11 = 1*9 + 2

9 = 4*2 + 1   =>  gcd(2311, 20) = 1

3 0
4 years ago
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