7. Solve for x.<br>V<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%3F%7D%20" id="TexFormula1" title=" \sqrt{?} " alt=" \sqr

At a point on the ground 46 feet from the foot of a tree, the angle of elevation of the top of the tree is 68 degrees. What is t

Mandarinka [93]

Answer:

114 ft

Step-by-step explanation:

Imagine or construct a right triangle with the 46 ft leg lying on the ground. This is the "adjacent side" of the triangle; it lies immediately adjacent to the 68 degree angle. The side opposite this angle is h, the height of the tree.

The tangent function includes angle, opp side and adj side:

tan 68 degrees = opp / adj = h / (46 ft), and so:

(46 ft)*tan (68 degrees) = opp = h

Then the height of the tree is h = (46 ft)(2.47) = 114 ft

6 0

3 years ago

The graph of y= (x - 2)(x + 4) is shown. What is the y-intercept of this graph?

LenKa [72]

Answer:

The y intercept in -8

Step-by-step explanation:

6 0

3 years ago

Billy drove 1 half of the 7.5 miles between his house and the mall. How many miles did he drive?

lara31 [8.8K]

Billy drove 3.75 miles. You find the answer by dividing 7.5 by 2

8 0

3 years ago

What is the area, in square inches, of this trapezoid?

Kisachek [45]

Answer:

60

Step-by-step explanation:

Area = ((a+b)/2)h=((7+13)/2)*6=60

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

7. Solve for x.V