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Degger [83]
4 years ago
11

Write seventeen thousand and one hundred six thousand in standard form

Mathematics
1 answer:
Zigmanuir [339]4 years ago
6 0
17000.106 is how you write the number above in standard form

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Write an<br> explicit formula for An, the nth term of the sequence 9, 16, 23, ....
gizmo_the_mogwai [7]
someone help meeeee
4 0
3 years ago
Read 2 more answers
Find the root(s) of f (x) = (x + 5)3(x - 9)2(x + 1). -5 with multiplicity 3 5 with multiplicity 3 -9
faust18 [17]

Answer:

  -5, multiplicity 3; +9, multiplicity 2; -1

Step-by-step explanation:

The roots of f(x) are those values of x that make the factors be zero. For a factor of x-a, the root is x=a, because a-a=0. If the factor appears n times, then the root has multiplicity n.

f(x) = (x+5)^3(x-9)^2(x+1) has roots ...

  • -5 with multiplicity 3
  • +9 with multiplicity 2
  • -1
4 0
3 years ago
Christov and Mateus gave out candy to children on Halloween. They each have out candy at a constant rate, and they both gave awa
padilas [110]

Answer: 1) Both gave same number of candies to each child.

2) Christov gave candy to a greater number of children.

Step-by-step explanation:

Since, Christov initially had 300 pieces of candy, and after he was visited by 17 children, he had 249 pieces left.

Let he gave x candy to each student when he visited 17 children.

Then, 17 x + 249 = 300

⇒ 17 x = 51

⇒ x = 3

Since, he distributes candies in a constant speed.

Therefore, his speed of distribution = 3 candies per child

Now,The function that shows the remaining number of candies Mateus has after distributing candies to n children,

C(n)=270 - 3 n

Initially, n = 0

C(0) = 270

⇒ Mateus has initial number of candies = 270

When he gave candies to one child then remaining candies = 270 - 3 × 1 = 267

Thus, the candies, get by a child from Mateus = 270 - 267 = 3

Since, he distributes candies in a constant speed.

⇒ His speed of distribution = 3 candies per child

1) Therefore, Both Christov and Mateus have same speed of distribution.

2) Since, both have same seed.

⇒ The one who has greater number of candies will be distribute more.

⇒ Christov will give more candies.


8 0
4 years ago
Suzi bought 5 packages of M&amp;M's, and each package contained 18 candy pieces. She opened the first package and counted 4 gree
Leni [432]

Answer: The correct answer is option B: There are between 15 and 20 green pieces in all 5 packages

Step-by-step explanation: The most important factor has been given which is, "Which statement about the candy pieces in the remaining packages is best supported by this information."

The information given is such that, the first package she opened had 4 green pieces and on this basis we can safely assume that all other packages have 4 green pieces as well. The second package had 3 green pieces and this based on this too we can safely assume that all other packages had 3 green pieces. Hence, all 5 packages can either have a total of 4 x 5 green candies which equals a total of 20 green pieces or, all 5 packages can have a total of 3 x 5 green candies which equals a total of 15 green pieces.

So according to Suzi's experiment, there are between 15 and 20 green pieces in all 5 packages.

7 0
4 years ago
What is [b-5] over 8=5
julia-pushkina [17]

Answer:

B=45

Step-by-step explanation:

Multiply 8 by 5

Add 40 and 5 together

6 0
3 years ago
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