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Arisa [49]
3 years ago
14

Is the orthocenter of a triangle on the inside ?

Mathematics
2 answers:
Olegator [25]3 years ago
8 0
It isn’t always in the triangle
likoan [24]3 years ago
7 0

Answer:

It turns out that all three altitudes always intersect at the same point - the so-called orthocenter of the triangle. The orthocenter is not always inside the triangle. If the triangle is obtuse, it will be outside. To make this happen the altitude lines have to be extended so they cross.

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Can someone please help me with these two questions! :( ill mark u as brainlest
Maslowich

Answer:

1.A by SSS axiom

2.F as angle A is 62*

4 0
3 years ago
Consider these functions:<br> f(x) = x + 1<br> g(x)= 2/x
kow [346]

Answer:

its letter A.

\frac{2}{x + 1}

3 0
3 years ago
What is the length of AC?
Alenkasestr [34]
So in order to find line AC you must find line AD and DC then plus them together.

to find AD use Pythagoras theorem
a^2 = c^2 - b^2
AD^2 = 7.5^2 - 6.5^2
AD^2 = 56.25 - 42.25
AD^2 = 14
square root both sides to get rid of the ^2
AD ≈ 3.7 or 3.74

Do the same for DC
DC^2 = 10^2 - 6.5^2
DC^2 = 100 - 42.25
DC^2 = 57.75
DC ≈ 7.6

now plus AD and DC which should give u 11.3
5 0
3 years ago
Find the Value of X (In this picture)​
slava [35]

Answer:

x = 63

Step-by-step explanation:

It's Quadrilateral so:

360-90-72= x + x +22 (Cause that 2 triangle is isosceles)

2x=126

x=63

4 0
3 years ago
Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering
Svetllana [295]

Answer:

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Step-by-step explanation:

We are asked to find the tangent line approximation for f(x)=\sqrt{10+x} near x=0.

We will use linear approximation formula for a tangent line L(x) of a function f(x) at x=a to solve our given problem.

L(x)=f(a)+f'(a)(x-a)

Let us find value of function at x=0 as:

f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}

Now, we will find derivative of given function as:

f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}

f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)

f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1

f'(x)=\frac{1}{2\sqrt{10+x}}

Let us find derivative at x=0

f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}

Upon substituting our given values in linear approximation formula, we will get:

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)  

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Therefore, our required tangent line for approximation would be L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x.

8 0
3 years ago
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