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Luba_88 [7]
2 years ago
10

Which statements describe the sequence –3, 5, –7, 9, –11, …? Check all that apply. The sequence has 5 terms. The 4th term of the

sequence is 9. f(5) = 2 The domain of the sequence is all natural numbers. (4, 9) lies on the graph of the sequence.
Mathematics
2 answers:
marysya [2.9K]2 years ago
6 0

Answer:

roof off God sell

Step-by-step explanation:

hahahahahah

Yuki888 [10]2 years ago
5 0

<u>forth term in a sequence =9</u>

<u>domain=set of all natural number</u>

<u>(4,9) lies in a graph</u>

Answer:

Solution given:

–3, 5, –7, 9, –11, …are in a sequence

domain=set of all natural number.

forth term in a sequence =9

(4,9) lies in a graph.

and

f(5)=2

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A football stadium sells regular and box seating. There are twelve times as many
Vera_Pavlovna [14]

Answer:

There are 801 box seats and 9612 regular seats.

explanation:

Let the number of box seats be x,

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The sum of seats equals to 10,413

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2 years ago
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56of the 90pets in the pet show are cats 4/5of the cats are calico cats. What fraction of the pets are calico cats?
meriva
To get the answer, we can divide 56 by 5 to find one fifth of the cats. Then, we can multiply this number to get four fifths of the cats. Then, we put this number over 90 since there is 90 pets in the entire show. Then, this answer ends up being about half all of the pets are calicos.
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3 years ago
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If E F = 7 , A C = 16 , and A F = 9
Vilka [71]

Answers:

CB = 14

GF = 8

FB = 9

EF is parallel to CB

====================================

Explanations:

Points E and F are midpoints of their respective sides. They form the midsegment EF. Because EF is a midsegment, A midsegment is half the length of its parallel counterpart, so CB is two times longer than EF. If EF is 7 units long, then CB = 2*EF = 2*7 = 14

For similar reasons, GF is parallel to AC. If AC = 16, then half of that is GF = (1/2)*AC = 0.5*16 = 8.

FB = FA = 9 as these segments have the same single tickmark to indicate they are the same length

EF is parallel to CB because EF is a midsegment, and this is one of the properties of being a midsegment. We can show that quadrilateral EGBF is a parallelogram to help prove this.

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3 years ago
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