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3 years ago

Someone please help me with this question asap.

Mathematics

1 answer:

vova2212 [387]3 years ago

7 0

Answer:

-1<x<1 and 1<x

Step-by-step explanation:

We are asked to determine the interval in which our function shown in the graph has positive values.

In order to do so, we have to see for what values of x on x axis, the graph is above x axis.

As we can see in the graph, when we move from x = -1 towards right, the graph is above x axis. And towards left of x=-1 , the graph is below x axis. Hence answer is

-1<x<1 and 1<x

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Please help with the questions in the image

Pavel [41]

Answer: f(x) = (x - 3)²

<u>Step-by-step explanation:</u>

Use the vertex formula: y = a(x - h)² + k

and input the vertex (h, k) = (3, 0) and the given point (x, y) = (4, 2) to solve for a

2 = a(4 - 3)² + 0

2 = a(1)²

2 = a

Now, input (h, k) = (3, 0) and a = 1 into the vertex formula:

y = 1(x - 3)² + 0 → y = (x - 3)²

****************************************************************************

Answer: g(x) = -(x + 1.25)² + 2.5625

<u>Step-by-step explanation:</u>

Use the vertex formula: y = a(x - h)² + k

and input the given points for (x, y) to create a system of equations, then solve for a, h, and k.

EQ1: 2 = a(-2 - h)² + k

2 = a(4 + 4h + h²) + k

2 - 4a - 4ah - ah² = k

EQ2: 1 = a(0 - h)² + k

1 = ah² + k

1 - ah² = k

EQ3: -2.5 = a(1 - h)² + k

-2.5 = a(1 - 2h + h²) + k

-2.5 -a + 2h + ah² = k

<u>Substitute</u> - set EQ1 = EQ2 and EQ2 = EQ3 to eliminate k

EQ1 = EQ2: 2 - 4a - 4ah - ah² = 1 - ah²

1 - 4a - 4ah = 0

EQ2 = EQ3: 1 - ah² = -2.5 - a + 2ah - ah²

3.5 + a - 2ah = 0

<u>Elimination:</u> - now solve the system for "a"

1 - 4a - 4ah = 0 → 1(1 - 4a - 4ah = 0) → 1 - 4a - 4ah = 0

3.5 + a - 2ah = 0 → -2(3.5 + a - 2ah = 0) → <u> -7 - 2a + 4ah = 0 </u>

-6 - 6a = 0

-6a = 6

Next, replace "a" with -1 into either of the equations to solve for "h"

1 - 4a - 4ah = 0

1 - 4(-1) - 4(-1)h = 0

1 + 4 + 4h = 0

5 + 4h = 0

4h = -5

Now, replace "a" with -1 and "h" with -1.25 into any of the original equations (EQ1, EQ2, or EQ3) to solve for k:

1 - ah² = k

1 - (-1)(-1.25)² = k

1 - (-1)(1.5625) = k

1 + 1.5625 = k

Now, input (h, k) = (-1.25, 2.5625) and a = -1 into the vertex formula:

y = -1(x - (-1.25))² + 2.5625 → y = -(x + 1.25)² + 2.5625

4 0

4 years ago

The graph of sinusidal function intersects its midline at (0,-3) and then has a maximum point at (2,-1.5). Write the formula of

Crank

Answer:

y = 1.5sin[(pi/4)x] - 3

Step-by-step explanation:

Mid line: y = -3

Amplitude: -1.5 - (-3) = 1.5

From mean line to max is ¼ of a cycle:

Period = 4 × 2 = 8

No. of cycles in 2pi:

2pi/8 = pi/4

y = 1.5sin[(pi/4)x] - 3

4 0

3 years ago

Read 2 more answers

Suppose that you began a one-year study of tuberculosis (TB) in a subsidized housing community in the Lower East Side of New Yor

Rufina [12.5K]

Answer:

The subsidized housing community in the lower East side of New York City is a dynamic population because people moved out of the population and others died while the survey was still being conducted.

The prevalence of TB on the screened community on January 1st is $\frac{30}{500} or = 0.06$

Step-by-step explanation:

Prevalence is given by; the number of persons affected by TB/ total number of people in the the study or survey.

3 0

3 years ago

A square carpet of side length 9 feet is designed with one large shaded square and eight smaller, congruent shaded squares, as s

Pepsi [2]

Answer:

$3\sqrt{3} + 72$

Step-by-step explanation:

I attached an image to aid the understanding of the question.

Looking at the image, we see that the 8 shaded parts are congruent, as affirmed in the question as well. And we are told that T = 3, this implies that the area of the square with T as it's side is 9ft². Since all the 8 squares are congruenrt, it means each square has its area to be 9. Therefore, the total area of the 8 shaded squares will be:

$9 \times 8 = 72ft^{2}$

It remains the area of the shaded square with side S.

From the question, we have the following ratio:

$\frac{9}{S} = \frac{S}{T}[\tex]But[tex]\frac{9}{S} = \frac{S}{3}[\tex]Multiplying through by 3S we have27 = S² and this gives:[tex]S = \sqrt{27} = 3\sqrt{3}$