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leva [86]
3 years ago
9

Is 2 a solution to 4x - 3​

Mathematics
1 answer:
AlexFokin [52]3 years ago
8 0

Answer:

2+4x-3=6

6-3=3

Answer:3

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Suppose that the function g is defined, for all real numbers, as follows.
svetoff [14.1K]

Answer:

<em>a) g(-2) = 0</em>

<em>b) g(-1) =1</em>

<em>c) g(4) = 2</em>

Step-by-step explanation:

Given data

g(x) = \frac{-1}{4x+1} if x < -2

g(x) = - (x+1)^{2} +1  if -2\leq x\leq 2

g(x) = 2 if x >2

<u><em>Step( i )</em></u>:-

g(x) = - (x+1)^{2} +1  if -2\leq x\leq 2

put x = -2

g(-2) = -(-2+1)^{2} +1 = -(-1)^{2}+1 = -1+1 =0

<em>g(-2) = 0</em>

<u><em>Step(ii)</em></u>:-

g(x) = - (x+1)^{2} +1  if -2\leq x\leq 2

Put x = -1

g(-1) = -(-1+1)^{2} +1 = -(0)^{2}+1 = -0+1 =1

<em>g(-1) =1</em>

<u><em>Step(iii)</em></u>:-

g(x) = 2 if x >2

<em>g(4) = 2</em>

<em></em>

7 0
3 years ago
Find the quotient. 8,489÷9
eimsori [14]
The answer is 943.2.
4 0
3 years ago
Read 2 more answers
What is the solution of n^2 - 49 = 0
Jlenok [28]
N² - 49 = 0
<u>   + 49  + 49</u>
        n² = 49
         n = <u>+</u>7

The solution to the problem is {7, -7}.
5 0
3 years ago
Read 2 more answers
300 ml of pure alcohol is poured from a bottle containing 2 l of pure alcohol. Then, 300 ml of water is added into the bottle. A
Ede4ka [16]

Answer:

The present percentage of pure alcohol in the solution is 72.25% of pure alcohol

Step-by-step explanation:

The volume of pure alcohol poured from the 2 l bottle of pure alcohol = 300 ml of pure alcohol

The volume of water added into the bottle after pouring out the pure alcohol = 300 ml of water

The volume of diluted alcohol poured out of the bottle = 300 ml of diluted alcohol

The volume of water added into the bottle of diluted alcohol after pouring out the 300 ml of diluted alcohol = 300 ml of water

Step 1

After pouring the 300 ml of pure alcohol and adding 300 ml of water to the bottle, the percentage concentration, C%₁ is given as follows;

C%₁ = (Volume of pure alcohol)/(Total volume of the solution) × 100

The volume of pure alcohol in the bottle = 2 l - 300 ml = 1,700 ml

The total volume of the solution = The volume of pure alcohol in the bottle +  The volume of water added = 1,700 ml + 300 ml = 2,000 ml = 2 l

∴ C%₁ = (1,700 ml)/(2,000 ml) × 100 = 85% percent alcohol

Step 2

After pouring out 300 ml diluted alcohol from the 2,000 ml, 85% alcohol and adding 300 ml of water, we have;

Volume of 85% alcohol = 2,000 ml - 300 ml = 1,700 ml

The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol = 85/100 × 1,700 = 1,445 ml

The total volume of the diluted solution = The volume of the 85% alcohol in the solution + The volume of water added

∴ The total volume of the twice diluted solution = 1,700 ml + 300 ml = 2,000 ml

The present percentage of pure alcohol in the solution, C%₂ = (The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol)/(The total volume of the diluted solution) × 100

∴ C%₂ = (1,445 ml)/(2,000 ml) × 100 = 72.25 %

The present percentage of pure alcohol in the solution, C%₂ = 72.25%

3 0
3 years ago
(e+3)(e-5) expanded and simplified
Zarrin [17]

Step-by-step explanation:

(e + 3)(e - 5) \\ e \times e + e \times  ( - 5) + e \times e + e \times 3 \\  { e}^{2}  + ( - 5e) +  {e}^{2}  + 3e \\  {2e}^{2}  + ( - 2e)

In this type of math equation, you have to multiply all members from the first part with the second part.

Expanded:

e \times e+e \times (-5)+e \times e+e \times 3

simplified:

{2e}^{2}  + ( - 2e)

8 0
2 years ago
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