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vovikov84 [41]
3 years ago
12

4 - t = 3(t - 1) - 5

Mathematics
1 answer:
Anika [276]3 years ago
4 0

Answer:

T=3

Step-by-step explanation:

4-T = 3(t-1) -5

4-t=3T -3-5

^^ distribute

-T=3T-12

^^combine like terms and subtraction prop. of equality.

-4T=-12

^^ subtraction property of equality

(-4T=12)/-4

^^ thats just getting rid of the -sign, then its also simpliying

T=3

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In the table below, y is a linear function of x.
motikmotik

Answer:

y= -7

Step-by-step explanation:

y=ax+b

Find a where a=y2-y1 /x2-x1

y2=13,y1=5 x2=5,x1=3

substituting the values

a=13-5/5-2

a=8/2

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substituting a into the formular

y=4x+b

taking x and y values let say 3 and 5 for x and y

the equation becomes

5=4(3)+b

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substituting the values of a and b into the equation

y=4x-7

when x=0

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y=0-7

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3 0
1 year ago
. Using the Binomial Theorem explicitly, give the 15th term in the expansion of (-2x + 1)^19
timofeeve [1]
Let's rewrite the binomial as:
(1 - 2x)^{19}

\text{Binomial expansion:} (1 + x)^{n} = \sum_{r = 0}^n\left(\begin{array}{ccc}n\\r\end{array}\right) (x)^{r}

Using the binomial expansion, we get:
\text{Binomial expansion: } (1 - 2x)^{19} = \sum_{r = 0}^{19}\left(\begin{array}{ccc}19\\r\end{array}\right) (-2x)^{r}

For the 15th term, we want the term where r is equal to 14, because of the fact that the first term starts when r = 0. Thus, for the 15th term, we need to include the 0th or the first term of the binomial expansion.

Thus, the fifteenth term is:
\text{Binomial expansion (15th term):} \left(\begin{array}{ccc}19\\14\end{array}\right) (-2x)^{14}
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3 years ago
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Answer:

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Answer:

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Step-by-step explanation:

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