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vovangra [49]
4 years ago
8

What would be an estimate for the number 726

Mathematics
2 answers:
DedPeter [7]4 years ago
8 0

730 or 725!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

kozerog [31]4 years ago
4 0
Answer is 730!!!!!!!
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.
babunello [35]
<span>From the problem, you can get this equation:
Jane is 2 years older than Sally
J=z+2
David is 4 times older than Jane.
D=4J

</span>Write two equivalent expressions that represent David’s age in terms of Sally’s age, z, and explain how the two expressions relate to one another.<span>Sally is z years old.
From the 2 equation above, you can get:
</span>D=4J
D=4(z+2)
D=4z+8

D-8=4z
(D-8)/4=z
The two equivalent expression would be D=4z+8 and (D-8)/4=z

The expression that more usefull would be D=4z+8 if you want to determine David age.
4 0
4 years ago
A survey is conducted on students taking a statistics class. several variables are measured in the survey. which of these variab
k0ka [10]
Is this multiply choice???  If so, could you tell us the choice???
5 0
3 years ago
Create a set of data that shows temperature highs for 10 days and satisfies each condition below:
coldgirl [10]

72*10=720 so all the numbers would need to add to 720

the median is 74 so you need to have both 75 and 76 in the set

the mode is 68 so that need to be in at least twice

and the range is 21 so the largest number-21=smallest number

57, 68, 68, 68, 75, 76, 76, 77, 77, 78

7 0
4 years ago
Simplify.
djverab [1.8K]
15m + 2n, first you have to pair the matching variables, 8m and 7m along with the sign in front of them. so 8m(+7m), you would then add those then do the same with the other, +4n(postive) and -2n, so subtract 2 from 4 and get 2. put them together and it’s 15m+2
5 0
3 years ago
A Cepheid variable star is a star whose brightness alternately increases and decreases. Suppose that Cephei Joe is a star for wh
4vir4ik [10]

Answer:

\dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi t}{5.3})

0.16

Step-by-step explanation:

According to the question the mathematical model should be

B(t)=2.9+0.35\sin(\dfrac{2\pi t}{5.3})

Differentiating with respect to time we get

B'(t)=0.35\cos(\dfrac{2\pi t}{5.3})\times(\dfrac{2\pi}{5.3})\\\Rightarrow B'(t)=\dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi t}{5.3})

So, the rate of change of brightness after t days is \dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi t}{5.3})

After 1 day means t=1

B'(1)=\dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi\times 1}{5.3})\\\Rightarrow B'(1)=0.16

The rate of increase after one day is 0.16.

8 0
3 years ago
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