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IrinaVladis [17]
2 years ago
9

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

6 0
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
4 0
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:

8 0
3 years ago
Read 2 more answers
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
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