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3 years ago

Which interval for the graphed function has a local minimum of 0?

Mathematics

2 answers:

koban [17]3 years ago

8 0

Answer:

{2,4} I know cause I JUST took the same test yesterday :) I guess I didin't help much huh

Wittaler [7]3 years ago

7 0

Answer:

Interval for the function has local minimum of 0 is [2, 4].

Step-by-step explanation:

We need to find the local minimum of 0 in the given function's graph.

In mathematics, local minimum is a point on a graph whose value is less than all other points near it.

See that, graph value 0 is lie on the x = 3 and in interval x = 2 to x = 4

So, final answer is :

Interval for the function has local minimum of 0 is [2, 4].

That's the final answer.

I hope is helps you.

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The straight line joining the points A(3,-5) and B(6,k) has a gradient of 4. Work out the value of k.

BaLLatris [955]

The straight line joining the points A(3,-5) and B(6,k) has a gradient of 4.

Gradient is the slope

So the slope of the line joining the points A(3,-5) and B(6,k) is 4

Slope of line joining two points = $\frac{y_2-y_1}{x_2-x_1}$

A(3,-5) and B(6,k) are (x1,y1) and (x2,y2)

slope = $\frac{y_2-y_1}{x_2-x_1}$

slope = $\frac{k-(-5)}{6-3}=\frac{k+5)}{3}$

We know slope =4

$\frac{k+5)}{3}=4$

Cross multiply and solve for k

k + 5 = 12

k = 7

The value of k = 7

7 0

3 years ago

Point B is on line segment \overline{AC}

MA_775_DIABLO [31]

I really can’t help you but I’ll wish look for you

3 0

3 years ago

I need help quickly please have clear explanation and answer written in brainly

Andrei [34K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a

Vitek1552 [10]

Answer:

$2ab - 6a + 5b - 15$

Step-by-step explanation:

Given

$2ab - 6a + 5b + \_$

Required

Fill in the gap to produce the product of linear expressions

$2ab - 6a + 5b + \_$

Split to 2

$(2ab - 6a) + (5b + \_)$

Factorize the first bracket

$2a(b - 3) + (5b + \_)$

Represent the _ with X

$2a(b - 3) + (5b + X)$

Factorize the second bracket

$2a(b - 3) + 5(b + \frac{X}{5})$

To result in a linear expression, then the following condition must be satisfied;

$b - 3 = b + \frac{X}{5}$

Subtract b from both sides

$b - b- 3 = b - b+ \frac{X}{5}$

$- 3 = \frac{X}{5}$

Multiply both sides by 5

$- 3 * 5 = \frac{X}{5} * 5$

$X = -15$

Substitute -15 for X in $2a(b - 3) + 5(b + \frac{X}{5})$

$2a(b - 3) + 5(b + \frac{-15}{5})$

$2a(b - 3) + 5(b - \frac{15}{5})$

$2a(b - 3) + 5(b - 3)$

$(2a + 5)(b - 3)$

The two linear expressions are $(2a+ 5)$ and $(b - 3)$

Their product will result in $2ab - 6a + 5b - 15$

<em>Hence, the constant is -15</em>

3 0

3 years ago

PLEASE HELP!! WILL GIVE BRAINLIEST!!!!

Arturiano [62]

Answer:

y=1

Step-by-step explanation:

Given function y=m(x)

The question is to find the y value where x equals -2 so the easiest way is to plus x=-2 in the graph, we can get y=1.

Hopefully this helps, if you have any questions please feel free to ask them in the comment section below, I'll be more than happy to answer!

5 0

3 years ago