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3 years ago

Please help!!!:( image is above ^

Mathematics

2 answers:

jekas [21]3 years ago

8 0

Answer:

<h2>the length of diagonal will be </h2><h2>√l^2+b^2+h^2</h2>

so, given length breadth and height are given respectively 12cm, 8cm, 14cm

so,

=√12^2+8^2+14^2

=√144+64+196

=√404

=20.009975124422 answer

r-ruslan [8.4K]3 years ago

3 0

We need to first find the diagonal that moves along the base.

a²+b²=c²

12²+8²=c²

c²=208

c=√208

Now we have find the length for the answer.

(√208)²+14²=c²

c²=208+196

c²=404

c= 20.10 cm approx.

answer: C

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xz_007 [3.2K]

Answer:

Step-by-step explanation:

When we reflect over x-axis then the x-coordinate doesn't change and the y-coordinate changes its sign

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3 years ago

The area of a triangle is 20 cm², and the

grandymaker [24]

Answer:

Heavenly, we need the forumula for the area of a triangle, do you know it?

Step-by-step explanation:

Area of a triangle is (1/2 ) * B*H

where B = base

H = height

then

A = 1/2* B * H

A is given to be 20

B = b

H = b+4

plug all those into the formula,

20 = (1/2)*b*(b+4)

use your algebra skillz :P

20 = 1/2 *( $b^{2}$ + 4b)

20*2 = $b^{2}$ + 4b

40 = $b^{2}$ + 4b

0 = $b^{2}$ + 4b-40

rewrite the above so it's in x format

$x^{2}$ + 4x -40 = 0

use the quadratic formula to find x , and for our problem, remember that x represents the base, or b, but don't mix it up with the b in the quadatic fomula, with the b of the triangle, which is the base, they are differnt "b"s :P

X = (b± $\sqrt{b^{2}-4*a*c }$ ) / 2*a

a = 1

b = 4

c = -40

x = (4± $\sqrt{4^{2}-4*1*(-40) }$ ) /2*1

x = ( 4± $\sqrt{16+160}$ ) / 2

x = (4± $\sqrt{176}$ ) / 2

x = (4±13.266499)/2

x= 2±6.633249

x = 2 + 6.633249 = 8.633

b= 8.633 cm

5 0

2 years ago

Which of the following explains how ΔBEI could be proven similar to ΔCEH using the AA similarity postulate? Quadrilateral ABDC,

maria [59]

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i got it right on the test

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