Which equation is represented by the graph below y = In x<br> y = In X+1<br> y=et<br> y = x + 1

the triangles below are congruent and their corresponding parts are marked. name all the corresponding congruent angles and side

GuDViN [60]

Answer:

See explanation

Step-by-step explanation:

The triangles below are congruent and their corresponding parts are marked.

a) Angles marked with one arc are congruent, so $\angle B\cong \angle Z.$

Angles marked with two arcs are congruent, so $\angle A\cong \angle Y.$

Angles marked with three arcs are congruent, so $\angle C\cong \angle X.$

b) Sides marked with one segment are congruent, so $\overline {AC}\cong \overline {YX}.$

Sides marked with two segments are congruent, so $\overline {BC}\cong \overline {ZX}.$

Sides marked with three segments are congruent, so $\overline {AB}\cong \overline {YZ}.$

c) Name teo congruent triangle following the congruence of angles:

$\triangle BCA\cong \triangle ZXY$

6 0

3 years ago

Please help and thank you

makvit [3.9K]

Answer:

f(x) should be reflected across the y-axis
f(x) should be horizontally translated

Step-by-step explanation:

We imagine Jenessa's elevation plots might look like those in the attachment. The red graph would be of f(x) when preparing for the race, and the green graph would be f(x) during the race.

The green graph is not a simple reflection or translation of the red graph, but is a combination of both. Thus, the two answer choices shown above are both applicable.

8 0

2 years ago

A building is 964 feet tall. Two people are standing on the ground directly west of the building. The

Lina20 [59]

Answer:

The distance from both of them = 1463.925 ft

Step-by-step explanation:

The building is 964 ft tall . 2 people are standing on the ground directly west of the building. The first person looks up at an angle of 62° to the top of the building while the second person did same at an angle of 26°. The distance between them can be computed below.

The illustration forms a right angle triangle . Using the SOHCAHTOA principle let us find the distance of the second person from the building

tan 26° = opposite/adjacent

tan 26° = 964/adjacent

adjacent tan 26° = 964

adjacent = 964/tan 26°

adjacent = 964/0.48773258856

adjacent = 1976.49290328 ft

The distance from the second person to the building = 1976.493 ft

Distance of the first person to the building

tan 62° = opposite/adjacent

tan 62° = 964/adjacent

adjacent tan 62° = 964

adjacent = 964/tan 62°

adjacent = 964/1.88072646535

adjacent = 512.567892122

distance from the first person to the building = 512.568 ft

The distance from both of them = 1976.493 ft - 512.568 ft = 1463.925 ft

5 0

3 years ago

Reflect the triangle in the mirror line.

vredina [299]

Answer:

I can't include a picture but it will be the same shape except it will be flipped over and will look a bit like differently angled packman with the back of the mouth apart the same distance, with the mouth "pointier" part of it one line away from the red line from the other side.... Oof sorry if this didn't help, I tried my best...

Step-by-step explanation:

From the "point" of the blue triangle, go diagonally SE 2 spaces. From there, draw a straight line to the right for 3 spaces. From there, draw upwards two spaces, and then connect those two lines into a triangle.

5 0

2 years ago

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

2 years ago

Which equation is represented by the graph below y = In x y = In X+1 y=et y = x + 1