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Afina-wow [57]
3 years ago
10

What kind of triangle must have an altitude that is outside the triangle?

Mathematics
2 answers:
Tanya [424]3 years ago
7 0

Answer:

Brainliest?

'''For an obtuse-angled triangle, the altitude is outside the triangle. For such triangles, the base is extended and then a perpendicular is drawn from the opposite vertex to the base. For an obtuse triangle, the altitude is shown in the triangle below.'''

Gekata [30.6K]3 years ago
6 0

Answer:

Obtuse

Step-by-step explanation:

When examining the shape of an obtuse triangle, due to it's wide nature, you will notice at least one of the side lengths will extend outward, which will create an altitude.

As outlined in this picture:

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What is 63 over 75 as a decimal?
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Answer:

0.84

Step-by-step explanation:

Use a calculator.  That would be 0.84.

                               3*21

Note that 63/75 = ---------- = 21/25.  Multiplying top and bottom by 4 results

                                3/25                   in 0.84 (so this answer checks out ok)

3 0
3 years ago
Select all complex numbers<br> 5-√9/4<br> 1+√-3<br> 4-3√-16<br> 2-√12➗5<br> 3+2√-9➗7<br> 9√-7/5
harina [27]

Answer:

5-√9/4    4-3√-16    3+2√-9➗7

Step-by-step explanation:


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3 years ago
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AlekseyPX
The slope is 2 and the y intercept is 4
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3 years ago
Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the
bekas [8.4K]

Answer:

Summation notation:

\frac{1}{2}\sum_{k=1}^6f((.5k))

or after using your function part:

\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

We want to divide that length of 3 units by 6 which gives a length of a half per each base length.

We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have  a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)

or by factoring out the 1/2 part:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

To find f(3) replace x in -x^2+2x+4 with 3:

-3^2+2(3)+4

-9+6+4

1

To find f(2.5) replace x in -x^2+2x+4 with 2.5:

-2.5^2+2(2.5)+4

-6.25+5+4

2.75

To find f(2) replace x in -x^2+2x+4 with 2:

-2^2+2(2)+4

-4+4+4

4

To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

4.75

To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

-1+2+4

5

To find f(.5) replace x in -x^2+2x+4 with .5:

-.5^2+2(.5)+4

-.25+1+4

4.75

Now let's add those heights.  After we obtain this sum we multiply by 1/2 and we have our approximate area:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

\frac{1}{2}(1+2.75+4+4.75+5+4.75)

\frac{1}{2}(22.25)

11.125

Okay now if you wanted the summation notation for:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

is it

\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))

or after simplifying a bit:

\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))

\frac{1}{2}\sum_{k=1}^6f((.5k))

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

6 0
3 years ago
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