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3 years ago

Jamie draws a triangle he says two of my three angles are obtuse

1 answer:

5 0

A triangle's interior angles will always add up to 180 degrees.

When one interior angle gets larger, the other angles must decrease in size.

If you attempt to make a triangle with two obtuse angles, you will find you can't do it in three lines. But you need an extra line in order to close the shape. But it won't be a triangle.

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Help ASAP????? Please

galben [10]

Answer:

C. (-5,-1)

E. (10,2)

F. (15,3)

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hope it helps...

have a great day!!

4 0

2 years ago

If A(3,4) and B(10,4) are the end points of a diameter of circle "O", what is the area of the circle O

ikadub [295]

Answer:

$\large\boxed{\dfrac{49\pi}{4}}$

Step-by-step explanation:

The formula of an area of a circle:

$A=\pi r^2$

r - radius

We have the endpoints of a diameter of a circle.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the given points A(3, 4) and B(10, 4):

$d=\sqrt{(10-3)^2+(4-4)^2}=\sqrt{7^2+0^2}=\sqrt{7^2}=7$

Therefore we have the length of the diameter: d = 7

We know: d = 2r. Therefore

$r=\dfrac{7}{2}$

Substitute to the formula of an area:

$A=\pi\left(\dfrac{7}{2}\right)^2=\dfrac{49\pi}{4}$

There is no such answer in the answers to choose.

Apparently you have some error in the coordinates of the endpoints.

8 0

2 years ago

WILL MARK BRAINLIEST what is the domain of (f/g) (x)?

tatyana61 [14]

Given that $f(x) = \sqrt{7-x}$ and $g(x) = \sqrt{x + 2}$ , we can say the following:

$\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).

Thus, let's set what is under both square roots to be greater than 0:

$\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7$

$\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2$

Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.

$x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}$

Now, let's look back at the function entirely, which is:

$\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Since $\sqrt{x + 2}$ is on the bottom of the fraction, we must say that $\sqrt{x + 2} \neq 0$ , since the denominator can't equal 0. Thus, we must exclude $\sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2$ from the domain.