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makkiz [27]
2 years ago
5

Kinda need help with 2 questions there short questions. Anyone....help

Mathematics
2 answers:
MatroZZZ [7]2 years ago
8 0

Answer:

70.56

Step-by-step explanation:

Darina [25.2K]2 years ago
7 0

Answer:

it seems the answer has already been figured out

Step-by-step explanation:

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Differentiation. questions 1 and 2
drek231 [11]
For 11, we start by plugging 1 in, getting 2(1)^2+1-5=-2=y

Plugging 1+m instead, we get 2(1+m)^2+1+m-5=2(1+2m+m^2)+1+m-5
= 2+4m+2m^2+1+m-5= 2m^2+5m-2. We have to find the difference between that and -2, so 2m^2+5m-2 -(-2)=2m^2+5m

Differentiating 12, we use the Power Rule to get -6x^2+14, and plugging x=1 into it, we get -6(1)^2+14=-6+14=18
4 0
3 years ago
What is an algebraic equation for four times a number y is sixteen
Len [333]

4y=16

y is the variable. If you were to solve the equation you would write the equation as 4(4)=16.

3 0
4 years ago
Use a distributive property to remove the parentheses -9(2v-u-3)
jeka57 [31]
-9(2v-u-3)
-18v+9u+27
4 0
3 years ago
Read 2 more answers
HELP ME PLZ AND IF YOU HELP ME 10 POINTS
mafiozo [28]

Answer:

3(c+4)

Step-by-step explanation:

It would usually be 3 x 4, but since there is a variable, it can`t be multiplied (unless you know what the value of the variable is). Thus the expression is 3(c+4).

6 0
3 years ago
Read 2 more answers
2. Margie Spencer wants to have $100,000 in her savings account in 20 years. If her account pays 6.6% annual
WITCHER [35]

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$100000\\ P=\textit{original amount deposited}\\ r=rate\to 6.6\%\to \frac{6.6}{100}\dotfill &0.066\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannually, thus two} \end{array}\dotfill &2\\ t=years\dotfill &20 \end{cases}

100000=P\left(1+\frac{0.066}{2}\right)^{2\cdot 20}\implies 100000=P(1.033)^{40} \\\\\\ \cfrac{100000}{1.033^{40}}=P\implies 27288.97\approx P

7 0
2 years ago
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