All categories

3 years ago

A triangle has side lengths 87cm 21cm and 93cm is the triangle acute obtuse or right

Mathematics

2 answers:

Lana71 [14]3 years ago

6 0

The triangle is obtuse.

Alex_Xolod [135]3 years ago

6 0

Answer: The given triangle is an obtuse triangle.

Step-by-step explanation: Let ABC be the given triangle, where the side lengths are as follows :

$AB=87~\textup{cm},~~BC=21~\textup{cm},~\textup{and}~CA=93~\textup{cm}.$

We are to check whether ΔABC is acute, obtuse or right.

We know that ΔABC will be

(i) Acute, if AB² + BC² > CA²,

(ii) Obtuse, if AB² + BC² < CA²,

(iii) Right, if AB² + BC² = CA².

Now, we have

$AB^2=87^2=7569,\\\\BC^2=21^2=441,\\\\CA^2=93^2=8649.$

So,

$AB^2+BC^2=7569+441=8010$

Therefore, the given triangle is an obtuse triangle.

You might be interested in

Use multiplying by 1 to find an expression equivalent to 3/14 with a denominator of 14 y.

Soloha48 [4]

As a matter of simplification, same/same = 1.

now, "same" could be anything whatsoever, you name it, a whole polynomial, a whole anything whatsoever.

$\bf \cfrac{\sqrt{2}}{\sqrt{2}}=\cfrac{\sqrt{3}}{\sqrt{3}}=\cfrac{100000000}{100000000}=\cfrac{eggs}{eggs}=\cfrac{miles}{miles}=\cfrac{\frac{\quad \sqrt{3}}{\sqrt[5]{17}}\quad }{\frac{\sqrt{3}}{\sqrt[5]{17}}}=\cfrac{whatever}{whatever}=1$

now, that said, let us use 1 in this case then ... hmmmm we simply need a "y" in the denominator, so we simply multiply the denominator AND numerator by "y", y/y = 1 btw.

$\bf \cfrac{3}{14}\cdot \cfrac{y}{y}\implies \cfrac{3y}{14y}$

6 0

4 years ago

At what point does she lose contact with the snowball and fly off at a tangent? That is

postnew [5]

Answer:

α ≥ 48.2°

Step-by-step explanation:

The complete question is given as follows:

" A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to the skier make with the vertical?"

- The figure is also attached.

Solution:

- The skier has a mass (m) and the snowball’s radius (r).

- Choose the center of the snowball to be the zero of gravitational potential. - We can look at the velocity (v) as a function of the angle (α) and find the specific α at which the skier lifts off and departs from the snowball.

- If we ignore snow-ski friction along with air resistance, then the one work producing force in this problem, gravity, is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical energy at the top of the snowball.

- Hence, From conservation of energy we have:

KE (α) + PE(α) = KE(α = 0) + PE(α = 0)

0.2*m*v(α)^2 + m*g*r*cos(α) = 0.5*m*[ v(α = 0)]^2 + m*g*r

0.2*m*v(α)^2 + m*g*r*cos(α) ≈ m*g*r

m*v(α)^2 / r = 2*m*g( 1 - cos(α) )

- The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity can hold her to that path, i.e. as long as:

m*g*cos(α) ≥ 2*m*g( 1 - cos(α) )

- Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal force—or else the skier will sink into the snowball.

- The expression for α_lift becomes:

3*cos(α) ≥ 2

α ≥ arc cos ( 2/3) ≥ 48.2°

4 0

3 years ago

Just one or four please help ..

eduard

For num 4 the answer to six is 22 the reason is this formula so write this down for n aswell
n(n+1)/2+1 the example could be as we know for 1 cut we always will have 2 slices so 1(1+1)/2+1 is 2 and for 6 cuts 6(6+1)/2+1 is 22 therefore the answer to 6 cuts is 22 and to n cuts is the formula n(n+1)/2+1

7 0

4 years ago

If two lines are parallel, then they do not intersect.

prisoha [69]

Answer:

Step-by-step explanation:

It clearly states that the lines are parallel. If they don't intersect that would be the answer. If you put B. you would only be repeating the information that was already given

8 0

3 years ago

Reflections. Please help me

Marat540 [252]

Answer:

see the attachments

Step-by-step explanation:

I suggested to you on a different occasion that by using tracing paper or tissue paper, you could make a copy of the image that you could move in the desired way to find its new location.

Here, the first attachment shows the figure being drawn on a piece of tissue with the line of reflection and the axes origin also shown.

The second attachment shows the tissue flopped over and the origin and line of reflection aligned with their previous locations. The new location of the figure is fairly obvious.

For <em>reflection</em>, any point that was some distance from the line on one side will be reflected to the same distance from the line on the other side. The distance is measured perpendicular to the line.

___

<em>Comment on "the work"</em>

I only have a badly focused image of your original worksheet to work from, but even that is sufficient to illustrate the process and the result. It took longer to make and edit the photos than to do the drawing necessary to find the answer.

8 0

3 years ago