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Alina [70]
3 years ago
7

The sides of a triangle are in the ratio 5 : 12 : 13. What is the length of each side of the triangle if the perimeter of the tr

iangle is 15 inches?
Mathematics
1 answer:
Ganezh [65]3 years ago
4 0

Answer:

2.5, 6 and 6.5 inches.

Step-by-step explanation:

5 + 12 + 13 = 30

So the shortest side = 5/30 * 15 = 1/6 * 15

= 2.5 inches.

The longest = 13 / 30 * 15 =  13/2

= 6.5 inches,

and the third = 12/30 * 15

= 6 inches.

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The coordinates of the vertices of △PQR are P(1, 4) , Q(2, 2) , and R(−2, 1) . The coordinates of the vertices of △P′Q′R′ are P′
sveticcg [70]

ANSWER


The correct  answer is B


<u>EXPLANATION</u>

If we analyse the coordinates carefully, you realize there is a mapping of

(x,y)\rightarrow(-x,y)

This is a reflection in the y-axis.


Since the transformation is a reflection, the shape is preserved. Therefore, the image triangle P'Q'R' of triangle

PQR are congruent.



5 0
3 years ago
Read 2 more answers
Find AA and BB that make the equation true. Verify your results.
CaHeK987 [17]

Answer:

a. A = -1 and B = 1

b. A = 7 and B = -5

Step-by-step explanation:

a.

\frac{A}{x+1} +\frac{B}{x-1}  = \frac{2}{x^2-1}

\frac{A*(x-1)+B*(x+1)}{(x+1)*(x-1)} = \frac{2}{x^2-1}

\frac{Ax - A + Bx + B}{x^2 -1} = \frac{2}{x^2-1}

To the equation be true, the "x-parts" and "nonx-parts" mist be the same, so:

Ax + Bx = 0

(A + B)x = 0

A + B = 0

A = -B

B - A = 2

B - (-B) = 2

2B = 2

B = 1  and A = -1

b.

\frac{A}{x+3} + \frac{B}{x +2} = \frac{2x -1}{x^2+5x+6}

\frac{A*(x+2) + B*(x+3)}{(x+3)*(x+2)} = \frac{2x-1}{x^2+5x+6}

\frac{Ax + 2A + Bx + 3B}{x^2 + 5x + 6} = \frac{2x-1}{x^2+5x+6}

To the equation be true, the "x-parts" and "nonx-parts" mist be the same, so:

Ax + Bx = 2x

(A + B)x = 2x

A + B = 2

A = 2 - B

2A + 3B = -1

2*(2-B) + 3B = -1

4 - 2B + 3B = -1

B = -5  and A = 2 - (-5) = 7

5 0
3 years ago
Find the equation of the line that passes through each pair of points in y = mx + b form.
mestny [16]

Answer:

y = 3x - 1

Step-by-step explanation:

6 0
3 years ago
Find 2 consecutive odd integers sum of who's square is 290​ pls send fast
ELEN [110]

Answer:

<h3>Hence required integers are 11 and 13</h3>

Let x an odd positive integer

Then, according to question

x^2 +(x+2)^2=290

2x^2 +4x−286=0

x^2 +2x−143=0

x ^2+13x−11x−143=0

(x+13)(x−11)=0

x=11 as x is positive

Hence required integers are 11 and 13

Step-by-step explanation:

Hope it is helpful....

7 0
3 years ago
Giving brainliest!
Y_Kistochka [10]
So using trigo ;

cos D =base/hypo
cosD =23.5/33.8 = 0.695

so and D = 45.97°

thus angle F = 180-(90+45.97) =44.03°

so your answers are 45.97° and 44.03° or you can say 45° and 45°....approx !!
7 0
3 years ago
Read 2 more answers
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