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tamaranim1 [39]
3 years ago
10

Is the expression 3(x+11/2)-3 equivalent to 3x+11/2​

Mathematics
2 answers:
Yanka [14]3 years ago
8 0

Answer:

No  

Step-by-step explanation:

3\left(x + \dfrac{11}{2}\right ) -3 = 3x + \dfrac{33}{2} -3 = 3x + \dfrac{33}{2} - \dfrac{6}{2} = 3x + \dfrac{27}{2}

Conclusion: The two expressions are not equivalent .

valentinak56 [21]3 years ago
3 0

Answer:i dont think so

Step-by-step explanation:

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I would really appreciate it if someone could answer these as i'm really stuc and its due for tomorrow.
k0ka [10]
1. Round 50.75 to 50 and 0.18 to 0.20.
50 x 0.2 = 10

2) Round 96 to 100 and 0.499 to 0.5
100 divided by 0.5 = 50

3a) Round 8.2 to 8, 6.7 to 7, and 0.46 to 0.50
8 x 7 divided by 0.50 = 112

3b) Round 23.4 to 20, 13.9 to 10, and 0.18 to 0.20
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7 0
3 years ago
How many solutions does the equation x^2-9x=-8 have?
Elza [17]
X^2-9x=-8
x^2-9x+8=-8+8
x^2-9x+8=0
(x-8)(x-1)=0

x-8=0
x-8+8=0+8
x=8

x-1=0
x-1+1=0+1
x=1

There are 2 solutions, which are 1 and 8 
8 0
3 years ago
Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t​
juin [17]

Answer:

a. 73; b. 48.9; c. 2; d. 33.8; e. 73

Step-by-step explanation:

Assume the function was

S(t)= 73 - 15 ln(t + 1), t  ≥ 0

a. Average score at t = 0

S(0) = 73 - 15 ln(0 + 1) = 73 - 15 ln(1) = 73 - 15(0) =73 - 0 = 73

b. Average score at t = 4

S(4) = 73 - 15 ln(4 + 1) = 73 - 15 ln(5) = 73 - 15(1.61) =73 - 24.14 = 48.9

c. Average score at t =24

S(24) = 73 - 15 ln(24 + 1) = 73 - 15 ln(25) = 73 - 15(3.22) =73 - 48.28 = 24.7

d. Percent of answers retained

At t = 0. the students retained 73 % of the answers.

At t = 24, they retained 24.7 % of the answers.

\text{Percent retention} = \dfrac{\text{24.7}}{\text{73}} \times 100 \, \% = \text{33.8 \%}\\\\\text{The students retained $\large \boxed{\mathbf{33.8 \, \%}}$ of their original knowledge after two years.}

e. Maximum of the function

The maximum of the function is at t= 0.

Max = 73 %

The graph below shows your knowledge decay curve. Knowledge decays rapidly at first but slows as time goes on.

 

6 0
3 years ago
Find an equation for the tangent to the curve at the given point y=x^3 , (2,8)
Yuliya22 [10]

\bf y=x^3\implies \left. \cfrac{dy}{dx}=3x^2 \right|_{x=2}\implies \stackrel{\stackrel{m}{\downarrow }}{12} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{8})~\hspace{10em} slope = m\implies 12 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-8=12(x-2) \\\\\\ y-8=12x-24\implies y=12x-16

5 0
3 years ago
Homework Question Help
harkovskaia [24]

Answer:

A=49 degrees

a= 41 degrees

b=90 degrees

Step-by-step explanation:

its asking for the degrees?

5 0
3 years ago
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