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Sonja [21]
1 year ago
14

Given the graph, which coordinate point is NOT a solution?

Mathematics
1 answer:
Natasha2012 [34]1 year ago
6 0

Answer: (1, 2) (The third option)

Step-by-step explanation: A solution has to be on the line. (1, 2) is the only point that isn't on the line.

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Can u please find the length of BC round to the nearest tenth
PolarNik [594]

Answer:

Step-by-step explanation:

Using sine rule

SinA/a = SinB/b = SinC/c

A = 84° a = BC, C = 51°, c = 16

Sin84/BC = Sin51/16

Cross multiply

BC Sin51 = 16 × Sin84

BC = 15.91/Sin51

BC = 20.48unit

5 0
3 years ago
has 8 times as many rocks in his collection as carter. carter and jacey have 54 rocks altogether. how many rocks does jacey have
lions [1.4K]

Answer:

Jacey has 48 rocks. Carter has 6 rocks.

If Carter and Jacey are the only factor owners of 54 rocks

We divide 48 by 8 = 6 rocks: 42 difference;

Which is also 54  as 6 + 48 = 54

They now have 48 difference

6,6, 6,6, 6,6, 6,6 = 48

We see 48 - 6 = 42.

We also see 48 + 6 = 54

54=6 = 48

48 difference is the denominating divider; meaning 48/8 = 6 is how we first find the answer before adding on the 6 to show the difference of 8 equal shares.

7 0
3 years ago
HELP HELP HELP HELP√ GIVES BRAINLIST ∵↓↑⇄⇆⇒≅₹
snow_tiger [21]
C I think I’m not sure
6 0
2 years ago
Please help if you can :)
mojhsa [17]

a) n² + 6

n=1

1² + 6 = 1 + 6 =7

n=2

2² + 6 = 4 + 6 = 10

n=3

3² + 6 = 9 + 6 =15

n=4

4² + 6 = 16 + 6 =22

7, 10, 15, 22

n=10

10² + 6 = 100 + 6 =106

10 th term-106

b) 3n²

n=10

3 x 10² = 3 x 100= 300

10th term- 300

Hope this helps!

6 0
2 years ago
If 150 grams of a radioactive isotope are present at 2:00 PM and 10 grams remain at 6:00 PM (the same day), what is the half-lif
pogonyaev

Initial amount, A_o=150\ g .

Final amount, A =10\ g .

Time taken, t = 6:00 - 2:00 = 4 hour.

We know,

A=A_o(\dfrac{1}{2})^{\dfrac{t}{h}}\\\\10 = 150 \times \dfrac{1}{2}^{\dfrac{4}{h}}\\\\2^{\dfrac{4}{h}}=15\\\\\dfrac{4}{h}= log_215\\\\h = \dfrac{4}{log_215}\\\\h = 1.024 \ hours

Therefore, the half-life of the isotope is 1.024 hours.

Hence, this is the required solution.

8 0
3 years ago
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