Change equation to slope intercept form 2x-3y=15

Solve for x.<br> 144°<br> 18 degrees<br> 30 degrees<br> 48 degrees<br> 120 degrees

disa [49]

Answer:

Step-by-step explanation:

you have a rotation angle of 360 °, take out the known values (360 ° -90 ° -144 ° = 126), 144 ° is part of a flat angle of 180 °, from 180 ° remove 144 and you have 36 °, they are opposite angles therefore the same. therefore 360 - 144 - 90 - 36 - 36 = 54.

54 is 3x, so x = 18 ° (54 : 3 = 18)

7 0

2 years ago

A box contains plain pencils and pens. A second box contains color pencils and crayons. One item from each box is chosen at rand

pochemuha

Answer:

1/2

Step-by-step explanation:

its 1 out of 2 chances

6 0

3 years ago

Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the sq

stiv31 [10]

Answer:

18

Step-by-step explanation:

Remark

This is one of those questions that can throw you. The problem is that do you include the original rectangle or not. The way it is written it sounds like you shouldn't

However if you don't the question gives you 2 complex answers. (answers with the sqrt( - 1) in them.

Solution

Let the width = x

Let the length = x + 5

Area of the rectangle: L * w = x * (x + 5)

Area of the smaller squares (there are 2)

Area = 2*s^2

x = s

Area = 2 * x^2

Area of the larger squares = 2 * (x+5)^2

Total Area

x*(x + 5) + 2x^2 + 2(x + 5)^2 = 120 Expand

x^2 + 5x + 2x^2 + 2(x^2 + 10x + 25) = 120 Remove the brackets

x^2 + 5x + 2x^2 + 2x^2 + 20x + 50 = 120 collect the like terms on the left

5x^2 + 25x + 50 = 120 Subtract 120 from both sides.

5x^2 + 25x - 70 = 0 Divide through by 5

x^2 + 5x - 14 = 0 Factor

(x + 7)(x - 2) = 0 x + 7 has no meaning

x - 2 = 0

x = 2

Perimeter

P = 2*w + 2*L

w = 2

L = 2 + 5

L = 7

P = 2*2 + 2 * 7

P = 4 + 14

P = 18

8 0

3 years ago

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 410 gram setting. It is

Delvig [45]

Answer:

i think it is 451 i believe

6 0

2 years ago

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 $\rm \triangle OBA$ :

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

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

3 years ago