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jeka94
3 years ago
14

PLZ HELP!! Which real numbers are zeroes of the function?

Mathematics
1 answer:
astra-53 [7]3 years ago
5 0

The correct answers are: 1/2, 2 and -2.


To get this, factor f(x)

f(x)=2x^3-x^2-8x+4

f(x)= (2x^3-x^2) - (8x -4)

f(x)= (2x-1)(x^2) + (2x-1)(-4)

f(x)= (2x-1)(x^2-4)

f(x)= (2x-1)(x-2)(x+2)


2x-1=0

2x=1

x=1/2


x-2=0

x=2


x+2=0

x=-2


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If c varies directly as (r + 1) and c = 8, when r= 3, calculate the value of r when c= 20.
Aneli [31]

Answer:

r = 9

Step-by-step explanation:

Given that c varies directly as (r + 1) then the equation relating them is

c = k(r + 1) ← k is the constant of variation

To find k use the condition c = 8 when r = 3, then

8 = k(3 + 1) = 4k ( divide both sides by 4 )

2 = k

c = 2(r + 1) ← equation of variation

When c = 20, then

20 = 2(r + 1) ← divide both sides by 2

10 = r + 1 ( subtract 1 from both sides )

9 = r

5 0
3 years ago
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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.

At x = -1 there is an asymptote. Asymptotes are things you get close to but don't get to. (Think of it as the "I'm Not Touching" game you play on car trips.) So we exclude -1.

At x = 4 there is an unfilled (or open) circle and that means we exclude 4.

At x = 5 there is a filled circle so we include 5.

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
4 years ago
Help please!!! AHAHSBHSBDR​
Degger [83]

Answer:

a = 8/29 thus: Step 1 is wrong!

Step-by-step explanation:

Solve for a:

8 - a/2 = 3 (4 - 5 a)

Hint: | Put the fractions in 8 - a/2 over a common denominator.

Put each term in 8 - a/2 over the common denominator 2: 8 - a/2 = 16/2 - a/2:

16/2 - a/2 = 3 (4 - 5 a)

Hint: | Combine 16/2 - a/2 into a single fraction.

16/2 - a/2 = (16 - a)/2:

(16 - a)/2 = 3 (4 - 5 a)

Hint: | Make (16 - a)/2 = 3 (4 - 5 a) simpler by multiplying both sides by a constant.

Multiply both sides by 2:

(2 (16 - a))/2 = 2×3 (4 - 5 a)

Hint: | Cancel common terms in the numerator and denominator of (2 (16 - a))/2.

(2 (16 - a))/2 = 2/2×(16 - a) = 16 - a:

16 - a = 2×3 (4 - 5 a)

Hint: | Multiply 2 and 3 together.

2×3 = 6:

16 - a = 6 (4 - 5 a)

Hint: | Write the linear polynomial on the left hand side in standard form.

Expand out terms of the right hand side:

16 - a = 24 - 30 a

Hint: | Move terms with a to the left hand side.

Add 30 a to both sides:

30 a - a + 16 = (30 a - 30 a) + 24

Hint: | Look for the difference of two identical terms.

30 a - 30 a = 0:

30 a - a + 16 = 24

Hint: | Group like terms in 30 a - a + 16.

Grouping like terms, 30 a - a + 16 = (-a + 30 a) + 16:

(-a + 30 a) + 16 = 24

Hint: | Combine like terms in 30 a - a.

30 a - a = 29 a:

29 a + 16 = 24

Hint: | Isolate terms with a to the left hand side.

Subtract 16 from both sides:

29 a + (16 - 16) = 24 - 16

Hint: | Look for the difference of two identical terms.

16 - 16 = 0:

29 a = 24 - 16

Hint: | Evaluate 24 - 16.

24 - 16 = 8:

29 a = 8

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 29 a = 8 by 29:

(29 a)/29 = 8/29

Hint: | Any nonzero number divided by itself is one.

29/29 = 1:

Answer: a = 8/29

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Answer:

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Step-by-step explanation:

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