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2 years ago

If the area of traingle TQR is 60 cm square,what is the area of the parallelogram PQRS?

Mathematics

2 answers:

eimsori [14]2 years ago

6 0

Answer:

my question was also same

yuradex [85]2 years ago

4 0

Answer:

my question was the same but nobody answered

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If (−4, 11) and (−6, 5) are the endpoints of a diameter of a circle, what is the equation of the circle?

lora16 [44]

Answer:

A. $(x+5)^2+(y-8)^2=100$

Step-by-step explanation:

1. The standard form of the equation of a circle is $(x-h)^{2} + (y-k)^2=r^2$

2. To find the center we require the midpoint of the 2 given points

center = $[\frac{1}{2}(-4-6),\frac{1}{2}(11+5)]$

=(-5,8)

3. The radius is the distance from the centre to either of the 2 given points calculate the radius using the distance formula

d= $\sqrt{(x_{2}+x_{1})^{2} +(y_{2}-y_{1})^{2}$

let $(x_{1},y_{1})= (-5,8) \\$ and $(x_2,y_2)=(-4,11)$

r= $\sqrt{(-4+5)^2+(11-8)^2} =\sqrt{81+9}=10$

--> $(x-(-5)^2+(y-8)^2=10^2$

--> $(x+5)^2+(y-8)^2=100$

4 0

3 years ago

Read 2 more answers

What is an equation of the line that passes through the point (2,−6) and is parallel to the line x−2y=8?

lorasvet [3.4K]

Answer:

y = x/2 - 7

Step-by-step explanation:

First, we need to find the slope of the given equation: x - 2y = 8

Subtract x from both sides

x - 2y = 8

- x - x

-2y = 8 - x

Divide both sides by -2

-2y/-2 = (8 - x)/-2

y = -4 + x/2

The slope of this equation is 1/2

So the equation of our parallel equation is y = x/2 + b

We have to find b, so plug in the given coordinates

-6 = 2/2 + b

-6 = 1 + b

Subtract 1 from both sides

-6 = 1 + b

- 1 - 1

b = -7

Plug it back into the original equation

y = x/2 - 7

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3 years ago

A relay race has 4 runners who run different parts of the race. There are 16

xz_007 [3.2K]

Your coach can select it 4 times I think

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3 years ago

Graph The function f(x) = |2x −4|

Aliun [14]

Answer:this is an absolute value function is there more to the question

Step-by-step explanation:

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3 years ago

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Find the x- and y- intercepts of parabola y=5x^2-16x+10

____ [38]

Y-INTERCEPT

$y = 5x^2 - 16x + 10$

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

$\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10$

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

$y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0$

We need to use the quadratic formula

The quadratic formula helps us find what values of $x$ make the equation = 0

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\$

The x-intercepts are at:

$(\frac{8 + \sqrt{14}}{5}, 0)\\(\frac{8 - \sqrt{14}}{5}, 0)$

5 0

2 years ago

If the area of traingle TQR is 60 cm square,what is the area of the parallelogram PQRS?​

If the area of traingle TQR is 60 cm square,what is the area of the parallelogram PQRS?