All categories

3 years ago

4) Which pair of triangles is congruent by Angle - Angle- Side

Mathematics

2 answers:

ruslelena [56]3 years ago

6 0

Answer:

Step-by-step explanation:

Two angles are congruent, and one side is congruent in the pair of triangles, option 2.

Mashcka [7]3 years ago

6 0

Answer:

2 should be the answer

Step-by-step explanation:

since two angles are equal and one side in equal you can write triangle ABC=TAY by angle angle side axiom

You might be interested in

1 2 3 4 5 6 7 8 9 10 TIME REMAINING 54:32 Because the lid of a marker is wider than the marker itself, a set of markers can be p

tia_tia [17]

Answer:

10 Square Inches

Step-by-step explanation:

Trapezoid

Base =12 inches

Height =10 inches

Top side= 10 inches.

Area of a Trapezoid $=\frac{1}{2}(a+b)h, $where a and b are the lengths of the base and top respectively$$

$=\frac{1}{2}(10+12)*10\\=5*22=110 \:Square\:Inches$

Area of the Trapezoid=110 Square Inches

Rectangle

Base = 12 inches

Height =10 inches.

Area of a Rectangle=Base X Height

=12 X 10

=120 Square Inches

Difference in Area between the two packages

Difference=Area of Rectangle-Area of Trapezoid

=120-110

=10 Square Inches

5 0

3 years ago

Read 2 more answers

??????????????????????????

Umnica [9.8K]

Solution:

Simple Interest = Principal Amount × Rate of Interest/100 × Time

Here, Principal Amount = $6000

Rate of Interest = 6%

Time = 4 years

Simple Interest = 6000 × 6/100 × 4 = $1440

3 0

3 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago

Plzzz help meee I am timed

Arisa [49]

Answer:

The relationsjip between the quantity is "x 30"

Step-by-step explanation:

multiply all the x values with 30, you will get the y values

8 0

3 years ago

Read 2 more answers

find the coordinates of the vertices of the triangle after a reflection across the parallel lines; first across the line x= -2 a

Kamila [148]

Answer:

Step-by-step explanation:

that the coordinates of the vertices of the triangle after a reflection across the parallel lines .

8 0

3 years ago