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3 years ago

5

The length of a rectangle is 5 yd less than double the width, and the area of the rectangle is 52 yd2 . find the dimensions of t

he rectangle.

1 answer:

charle [14.2K]3 years ago

3 0

The dimensions are: 4 x 13

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ABCD is a trapezoid.<br> What is the area of trapezoid?

lana [24]

Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h

The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²

The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h

Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²

Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²

6 0

3 years ago

Transversal cuts parallel lines and at points X and Y respectively. The point nearest to C is X, and the point nearest to D is Y

hammer [34]

Answer: The answer is (B) ∠SYD.

Step-by-step explanation: As mentioned in the question, two parallel lines PQ and RS are drawn in the attached figure. The transversal CD cut the lines PQ and RS at the points X and Y respectively.

We are given four angles, out of which one should be chosen which is congruent to ∠CXP.

The angles lying on opposite sides of the transversal and outside the two parallel lines are called alternate exterior angles.

For example, in the figure attached, ∠CXP, ∠SYD and ∠CXQ, ∠RYD are pairs of alternate exterior angles.

Now, the theorem of alternate exterior angles states that if the two lines are parallel having a transversal, then alternate exterior angles are congruent to each other.

Thus, we have

∠CXP ≅ ∠SYD.

So, option (B) is correct.

5 0

3 years ago

Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

1 year ago

Graph the following inequality. Then click to show the correct graph.<br><br><br>3x - 2y ≤ 6

max2010maxim [7]

Answer:

We want to graph the inequality:

3x - 2y ≤ 6

The first step is to write this as a linear equation, to do it, we can isolate y in one side of the inequality.

3x ≤ 6 + 2y

3x - 6 ≤ 2y

(3/2)x - 6/2 ≤ y

(3/2)x - 3 ≤ y

or:

y ≥ (3/2)x - 3

Because we have the symbol ≥

The points on the line are solutions, then the first part is to graph the line:

y = (3/2)*x - 3

Next, we have:

y equal to or larger than (3/2)*x - 3

Then we need to shade all the region above that line.

The graph can be seen below.

3 0

3 years ago

Ill give brainliest to correct answer!!!!!!!

algol13

12 feet squared or 3 feet squared. hopefully this helps i got 2 different answers

4 0

3 years ago