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Valentin [98]
3 years ago
13

A sail is in the form of a right triangle that is four times as high as it is wide.The sail is made from 32 square meters of mat

erial. What is its height?
Mathematics
1 answer:
Veronika [31]3 years ago
7 0
5x/2=32

5x=64

x=12.8 (width)

4x=51.2 (length)
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A company loses $5,400 as the result of a manufacturing defect. The 8 owners of the company must share the loss equally. What is
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Change in profit = x - (5,400/8)
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Answer:

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Given segments AB and CD intersect at E.
nata0808 [166]

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

  • \mathbf{AB = 3\sqrt{2}}
  • AB and CD are not congruent
  • AB does not bisect CD
  • CD does not bisect AB

<u>(a) Length of AB</u>

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}}

<u>(b) Are AB and CD congruent</u>

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}}

\mathbf{CD = 3}

By comparison

\mathbf{CD \ne AB}

Hence, AB and CD are not congruent

<u>(c) AB bisects CD or not?</u>

If AB bisects CD, then:

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

The above equation is not true, because:

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

Hence, AB does not bisect CD

<u>(d) CD bisects AB or not?</u>

If CD bisects AB, then:

\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:

brainly.com/question/20837270

7 0
2 years ago
Remember to check for extraneous solutions
larisa86 [58]

Domain:v\neq0\\\\\dfrac{2}{v}+\dfrac{2}{3}=\dfrac{v-6}{3v}\\\\\dfrac{2\cdot3}{v\cdot3}+\dfrac{2\cdot v}{3\cdot v}=\dfrac{v-6}{3v}\\\\\dfrac{6}{3v}+\dfrac{2v}{3v}=\dfrac{v-6}{3v}\\\\\dfrac{6+2v}{3v}=\dfrac{v-6}{3v}\iff6+2v=v-6\qquad\text{subtract 6 from both sides}\\\\2v=v-12\qquad\text{subtract v from both sides}\\\\\boxed{v=-12}\to\boxed{A.}

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3 years ago
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