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3 years ago

Which theorem can you use to show TVR is congruent SRV? SAS, AAS, ASA or SSS?

Mathematics

1 answer:

GalinKa [24]3 years ago

7 0

Answer:

ASA RULE

Step-by-step explanation:

As 2 angles are given and RV is common in both trianlges hence TVR is congruent to SRV by ASA rule

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A total of 505 tickets were sold for a school play they were either adult tickets or student tickets they were 55 more student t

arsen [322]

Hello There!

<u><em>The number of adult tickets sold in the school play = 225</em></u>

<u><em>Total Number Of Tickets = 505</em></u>

<u><em>Number Of Student Tickets = x + 55</em></u>

Number of adult tickets sold can be represented by X

Number of student tickets = X + 55

X+X+55=505

2x+55=505

2x=505-55

2x=450

X=225

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3 years ago

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3 years ago

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What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 27 − x^2? show your

LUCKY_DIMON [66]

Answer:

Option B.

Step-by-step explanation:

The given curve is

$y=27-x^2$

We need to find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve $y=27-x^2$ .

Let the vertex in quadrant I be (x,y), then the vertex in quadrant II is (-x,y) .

Length of the rectangle = 2x

Width of the rectangle = y

Area of a rectangle is

$Area=Length\times width$

$Area=2x\times y$

Substitute the value of y from the given equation.

$Area=2x(27-x^2)$

$A=54x-2x^3$ .... (1)

Differentiate with respect to x.

$\frac{dA}{dx}=54-6x^2$

Equate $\frac{dA}{dx}=0$ , to find the critical points.

$0=54-6x^2$

$6x^2=54$

Divide both sides by 6.

$x^2=9$

$x=\pm 3$

The value of x can not be negative because side length can not be negative.

Substitute x=3 in equation (1).

$A=54(3)-2(3)^3$

$A=162-54$

$A=108$

The area of the largest rectangle is 108 square units.

Therefore, the correct option is B.

7 0

3 years ago

Is each line parallel, perpendicular, or neither parallel nor perpendicular to the line −2x+3y=12?

givi [52]

We are given first equation: −2x+3y=12.

Converting it in slope-intercept form

3y=2x + 12

y= 2/3 x + 12/3

y= 2/3 x + 4.

<em>Slope of the given equation is 2/3.</em>

<em>Note: When slopes are same, lines would be parallel.</em>

<em>When slope are negative reciprocals, lines would be perpendicular.</em>

<em>When neither same nor negative reciprocal, the lines neither parallel nor perpendicular.</em>

1) First option : -2x+y=12.

In slope-intercept form y = 2x +12.

Slope is 2 there.

<h3>Neither parallel nor perpendicular to the line −2x+3y=12.</h3>

2) 3x+2y=-2

In slope-intercept form y = -3/2 x - 1.

Slope = -3/2.

Slope are negative reciprocal to slope of −2x+3y=12 equation.

<h3>Therefore, lines are perpendicular.</h3>

3) y=2/3x-1

Slope = 2/3.

<h3>Slopes are same therefore lines are parallel.</h3>

4) -2x+3y=11

y = 2/3 x + 11/3.

<h3>Slopes are same therefore lines are parallel.</h3>

4 0

4 years ago

2. The width of a rectangle is 12 inches less than its length. The perimeter of the rect-

Alona [7]

Answer:

$P= 2*Lenght + 2*Width$

Since the perimeter is 56 inches we can solve for the lenght with this equation:

$56 in = 2*12in + 2*Length$

And solving for the length we got:

$Length = \frac{56in -24 in}{2} 16 in$

So then the lenght = 16 inhes and the width of 12 inches

Step-by-step explanation:

For a rectangle of width 12 inches and lenght y inches we know that the perimeter is given by:

$P= 2*Lenght + 2*Width$

Since the perimeter is 56 inches we can solve for the lenght with this equation:

$56 in = 2*12in + 2*Length$

And solving for the length we got:

$Length = \frac{56in -24 in}{2} 16 in$

So then the lenght = 16 inhes and the width of 12 inches

5 0

3 years ago