Symmetric property of congruence:if BC =NP, then

Find k such that f(x) is continuous at x=30

Maru [420]

Answer:

this is correct

Step-by-step explanation:

f continous in x=30 means that it should be the same value for 1.25 x 30 and 1.10 x 30 + k so k =4.5

5 0

3 years ago

An online camping supplies outlet charges $14 for a canteen,

Vedmedyk [2.9K]

Answer:

Part A: 14x + 2

Part B: CJ's order is $30, and Cameron's order is $58.

Step-by-step explanation:

Part A: The total cost of the order would be $14 multiplied by the amount of canteens, plus the shipping price of $2. This is what my expression represents.

Part B:

CJ's Order:

14(2) + 2 =

28 + 2

$30

Cameron's Order:

14(4) + 2 =

56 + 2 =

$58

Hope this helps! :D

3 0

3 years ago

What is m∠A ? Please see attachment (The Triangle Sum Theorem)

goblinko [34]

The triangle sum theorem says all 3 interior angles must equal 180 degrees.

so, 43 + 35 = 78. [ 180 - 78 = 102 degrees for the missing angles shared by the 2 triangles.

102 + 18 = 120. [180 - 120 = 60}

leaving the measure of Angle A as 60 degrees.

5 0

4 years ago

Read 2 more answers

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .