(-a)^3 in expanded form?

What’s the answer? Because am having a hard time with this math problem.

Studentka2010 [4]

9514 1404 393

Answer:

34.5 square meters

Step-by-step explanation:

We assume you want to find the area of the shaded region. (The actual question is not visible here.)

The area of the triangle (including the rectangle) is given by the formula ...

A = 1/2bh

The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...

A = (1/2)(11 m)(9 m) = 49.5 m^2

The rectangle area is the product of its length and width:

A = LW

The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...

A = (5 m)(3 m) = 15 m^2

The shaded area is the difference between the triangle area and the rectangle area:

shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2

The shaded region has an area of 34.5 square meters.

7 0

3 years ago

What are the ordered pairs of the

OlgaM077 [116]

Answer:

The two points solutions to the system of equations are: (2, 3) and (-1,6)

Step-by-step explanation:

These system of equations consists of a parabola and a line. We need to find the points at which they intersect:

$x^2-2x+3=-x+5\\x^2-2x+x+3-5=0\\x^2-x-2=0\\(x-2)(x+1)=0$

Since we were able to factor out the quadratic expression, we can say that the x-values solution of the system are:

x = 2 and x = -1

Now, the associated y values we can get using either of the original equations for the system. We pick to use the linear equation for example:

when x = 2 then $y=-(2)+5=3$

when x= -1 then $y=-(-1)+5=6$

Then the two points solutions to the system of equations are: (2, 3) and (-1,6)

6 0

3 years ago

Can monkeys eat pee when in well that has wet stuff in a zoo with harambe

shtirl [24]

thats funny!! love that!!!! please mark me

8 0

2 years ago

A certain academic program boasts that 87% of their graduates find full-time employment in their field within the first year of

WARRIOR [948]

Answer:

Option A, there is not sufficient sample evidence to conclude that the full-time placement rate is now less than 87% because the p-value is greater than 0.05.

Step-by-step explanation:

Here the Null hypothesis would be

H0: 87% of the graduates find full-time employment in their field within the first year of graduation

H1: Less than 87% of the graduates find full-time employment in their field within the first year of graduation

Here the p values is 0.07.

Since the p value is greater than 0.05, there are not enough evidences to reject the hull hypothesis.

Hence, option A is correct

7 0

3 years ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution

6 0

3 years ago