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patriot [66]
3 years ago
10

Which equations are equivalent to ? 3^3√x+4 =12 Check all that apply.

Mathematics
2 answers:
vova2212 [387]3 years ago
8 0

Answer:

The third and the fifth equations are equivalent.

Step-by-step explanation:

First of all, you must find x from 3\cdot \sqrt[3]{x+4}=12 and check during the procedure what expressions from the options are equivalent.

The process is shown below:

  1. 3\cdot \sqrt[3]{x+4}=12
  2. \sqrt[3]{x+4}=\frac{12}{3}
  3. \sqrt[3]{x+4}=4
  4. (\sqrt[3]{x+4})^3=(4)^3
  5. x+4=64
  6. x=64-4
  7. x=60

From step 3, it could be said that the first, the second and the fourth options ARE NOT equivalent equations because when the left side is \sqrt[3]{x+4}, the right side must be +4, and these three options don't meet the condition.

From step 5, it could be said that the third option IS an equivalent equation because it is the same expression.

From step 7, it could be said that the fifth option IS an equivalent equation because it is the same expression.

Thus, the third and the fifth equations are equivalent.

eimsori [14]3 years ago
4 0

Answer:

option c

x + 4 = 64

and

option e

x = 60  

Step-by-step explanation:

Given in the question an expression

3∛x+4 = 12

∛x + 4 = 12/3

∛x + 4 = 4

Take cube on both sides of the equation

∛(x + 4)³ = 4³

x + 4 = 64

x = 64 - 4

x = 60  

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3 years ago
Question is in the picture<br>please help
LiRa [457]

When given the graph of a function, the domain would include all the points that there is a graph. The strategy is to find what <em>is not</em> included.

What we are looking for are points of discontinuity. Think of it as when you remove your pencil from the paper.

From left to right, the graph stops at x = -3. So anything less than -3 is in the domain. Next, the graph starts up again at x =-1 after an asymptote (the vertical dashed lines). This piece goes to x = 4. So our domain is from -1 to 4.

Lastly, there's a jump from 4 to 5 and the graph goes on again. After 5, we take all the stuff more than it. So x > 5 is in the domain.

So x < -3, - 1 < x < 4, and x > 5 appears to be our domain. However, end points needed to be checked to see if we include them or not. Again we go left to right.

At x = -3 there is a filled (or closed) circle and that means we include -3.

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Now we refine our domain for the endpoints.

x ≤ -3, -1 < x < 4, x ≥ 5 is our domain.

The problem gives us intervals, and we gave it in inequalities. When we include an endpoint we use brackets - [ and } and when we exclude and endpoint we use parentheses - ( and ). Let's go back to x ≤ -3. Anything less works, and -3 is included (closed circle). That interval is (-∞, -3]. Next is the piece between -1 and 4. Since both are excluded, (-1,4) is our interval. We include 5 to write x ≥5 as the interval [5,∞).

Put the bolded ones all together and use the union, ∪, symbol to connect them, since something on the graph could be in any piece.

Our domain is (-∞, -3] ∪(-1,4) ∪ [5,∞).

5 0
3 years ago
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