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3 years ago

How do I identify the congruent triangles in each figure

Mathematics

1 answer:

Virty [35]3 years ago

5 0

Congruent triangles have the exact same three sides and the exact same angles. They are identical. So if you look at your triangles carefully, they are congruent. :)

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A police officer wears a belt that carries an extra 10 pounds on their waistline. The police officer weighs p pounds.

Lelechka [254]

Answer: T = 10p

Step-by-step explanation: Since we don't know what is the total weight, we put the an variable, T to represent the total weight. The 10p means that the poilce officer gained 10 Pounds.

3 0

3 years ago

A translation maps (x, y) to

Troyanec [42]

Answer:

second quadrant

Step-by-step explanation:

(x, y ) → (x - 5, y + 3 ) means subtract 3 from the original x- coordinate and add 3 to the original y- coordinate.

(- 3, - 2 ) ← in third quadrant

→ (- 3 - 5, - 2 + 3) → (- 8, 1 ) ← in second quadrant

6 0

3 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

Points (0,0) and (2,1):

y = $\frac{1-0}{2-0}(x-0)$

y = $\frac{x}{2}$

Points (2,1) and (0,3):

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

Mass of the lamina that occupies region D is 4.

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$