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3 years ago

HELP FAST LOL HAHAHAH

Mathematics

1 answer:

OleMash [197]3 years ago

4 0

Answer:

184

Step-by-step explanation:

First, note that the sides that are opposite to each other are congruent (this will come into play in a second). We can find the area of each triangle by doing 0.5(b)(h) (aka half * base* height). Since we've established tha opposite sides are congruent and the triangles are opposite to each other, they must be congruent. That means that the area is going to be the exact same for both triangles. So, we get the expression (0.5)(6)(4)(2), which comes out to 6*4 = 24. Then, to find the area of the 2 rectangles that are the sides, just do 5*10*2 = 100 square meters for both rectangles combines. Lastly, we need to find the area of the floor. The dimensions of the floor are 6 and 10, so we just do 6*10 = 60. Now, add all those numbers together. We get 100+60+24 = 184. Hope this helps! Let me know if anything is still confusing.

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constant of variation = 3

Step-by-step explanation:

we know that b varies jointly with c and d

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b∝c∝d

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$b=k\frac{cd}{e}$

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substituting this in our equation for b:

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The length of a segment is the distance between its endpoints.

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(a) Length of AB

We have:

$\mathbf{A = (1,2)}$

$\mathbf{B = (4,5)}$

The length of AB is calculated using the following distance formula

$\mathbf{AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

So, we have:

$\mathbf{AB = \sqrt{(1 - 4)^2 + (2 - 5)^2}}$

$\mathbf{AB = \sqrt{18}}$

Simplify

$\mathbf{AB = 3\sqrt{2}}$

(b) Are AB and CD congruent

First, we calculate the length of CD using:

$\mathbf{CD = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

Where:

$\mathbf{C = (2, 4)}$

$\mathbf{D = (2, 1)}$

So, we have:

$\mathbf{CD = \sqrt{(2 -2)^2 + (4 - 1)^2}}$

$\mathbf{CD = \sqrt{9}}$

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By comparison

$\mathbf{CD \ne AB}$

Hence, AB and CD are not congruent

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If AB bisects CD, then:

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Hence, AB does not bisect CD

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$\mathbf{CD = \frac 12 \times AB}$

The above equation is not true, because:

$\mathbf{3 \ne \frac 12 \times 3\sqrt 2}$

Hence, CD does not bisect AB

Read more about lengths and bisections at:

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