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3 years ago

Points D and E are midpoints of the sides of triangle ABC. The perimeter of the triangle is 48 units.

Mathematics

2 answers:

MatroZZZ [7]3 years ago

5 0

The value of t = 2. <u>Option a)</u> 2 is the correct answer.

<u>Step-by-step explanation:</u>

Perimeter is the sum of three sides of a triangle.

Given that, the points D and E are midpoints of the sides.

Therefore,

side AB= AD+DB

AB= 3t + 3t = 6t

side BC= BE+EC

BC= 4t + 4t = 8t

side AC= 7t+6

Perimeter of the triangle= sum of (side AB+side BC+side AC)

48 = 6t+8t+7t+6

48 = 21t+6

t= 42/21

t= 2

zaharov [31]3 years ago

3 0

Answer:

Step-by-step explanation:

I just took the test

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Mademuasel [1]

Answer:

The average number of daily miles traveled by these individuals was 11.72 miles per day.

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3 years ago

The area of a rectangle is 490490 in squaredin2. the ratio of the length to the width is 5 : 25:2. find the length and the width

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3 years ago

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Step-by-step explanation:

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3 years ago

Percentages What percentage is R10 of R200? R10 100 X R200 <br>

Andrej [43]

R10 is 5% of R200

<h3>What percentage is R10 of R200?</h3>

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2 years ago

Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed as:

$y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}$

Thus, the area of the surface is as follows:

$\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA$

$= \iint_D \sqrt{x^2 +y^2 +1 } \ dA$

$= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta$

$=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr$

$= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0$

$= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]$

$= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]$

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2 years ago