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2 years ago

Similarity - Item 33751 Select the two figures that are similar to each other. help pls

Mathematics

1 answer:

earnstyle [38]2 years ago

7 0

Answer:

<em>Similar: First two shapes only</em>

Step-by-step explanation:

<u>Triangle Similarity Theorems </u>

There are three triangle similarity theorems that specify under which conditions triangles are similar:

If two of the angles are congruent, the third angle is also congruent and the triangles are similar (AA theorem).

If the three sides are in the same proportion, the triangles are similar (SSS theorem).

If two sides are in the same proportion and the included angle is equal, the triangles are similar (SAS theorem).

The first pair of shapes are triangles that are both equilateral and therefore have all of its interior angles of 60°. The AAA theorem is valid and the triangles are similar.

The second pair of shapes are parallelograms. The lengths are in the proportion 6/4=1,5 and the widths are in proportion 3/2=1.5, thus the shapes are also similar.

The third pair of shapes are triangles whose interior acute angles are not congruent. These triangles are not similar

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3 years ago

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Volume is a three-dimensional scalar quantity. The number of kleenex boxes that can fit in the large box is 4096.

<h3>What is volume?</h3>

A volume is a scalar number that expresses the amount of three-dimensional space enclosed by a closed surface.

Given the larger box have a length of the side as 4 feet. Also, the kleenex box is a cube with side lengths of 3 inches. Therefore, the number of cubes that will fit is,

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2 years ago

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zhenek [66]

Answer:

Step-by-step explanation:

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3 years ago

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

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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 $x_2$ .

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3 years ago

Similarity - Item 33751 Select the two figures that are similar to each other. help pls​

Similarity - Item 33751 Select the two figures that are similar to each other. help pls