What is the area of the polygon

I need help with this one. It’s #6. :((

jonny [76]

Answer:

C

Step-by-step explanation:

f(x) = 2ˣ + 1

-f(x) = -(2ˣ) − 1

First, let's find the y-intercept.

-f(0) = -(2⁰) − 1 = -2

Only C can be correct.

6 0

3 years ago

Hacer una representación gráfica para nuestro caso tenemos por un punto M pasan infinitas rectas?

vladimir1956 [14]

Answer:

make a graphical representation for our case do we have infinite lines pass through a point M?

Step-by-step explanation:

If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations. If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

5 0

3 years ago

What is the decimal equivalent of 11 80 ?

dezoksy [38]

Answer:

0.1375

Step-by-step explanation:

6 0

3 years ago

PLS HELP ME I ONLY HAVE # HOURS TO COMPLETE I WILL MARK THE BRAINLIEST!!!

nata0808 [166]

triangle on right is the same as the triangle on the left

use pythagorean theorem

hypotenuse will give to the sides of the rectangle

a^2 + b^2 = hypotenuse squared

10.4^2 = 108.16

15.3^2 = 234.09

342.25 = hypotenuse squared

take the square root in both sides

hypotenuse = the square root of 342.25 =

18.5

add up the areas of the 2 triangles and rectangle

triangle area is 1/2 times 10.4 times 15.3 =

79.56

2 triangles areas are 159.12

rectangle area is 18.5 × 7 = 129.5

159.12 + 129.5 = 288.62

answer 1 = 288.62

second question:

to get 1 side take the

square root of 702.25 which is

26.5

to get the perimeter

multiply 26.5 by 4 which is

106

answer 2 is choice B 106

3 0

2 years ago

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3>Question 1</h3>

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3>Question 2</h3>

The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x+2=0 \implies x=-2$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is negative is:

Solution: -2 < x < 4
Interval notation: (-2, 4)

3 0

1 year ago