Ask question

Ask question

All categories

2 years ago

9

Are they unique more than one or no triangle

1 answer:

Tanya [424]2 years ago

8 0

The measures of the angles of a triangle add to 180.

45 + 45 + 90 = 180
The measures of these three angles do add to 180, ,so there is at least one triangle with these angle measures.
Using AA triangle similarity, any triangle with the same angle measures will be similar.

From the figure you already see two triangles with angles 45-45-90. There is an infinite number of triangles with those angle measures.

You might be interested in

Help? With my math please. Will mark brainlist URGENT

vaieri [72.5K]

Answer:

1 -51

2 m and n, and a and b

Step-by-step explanation:

7 0

2 years ago

PLEASE HELP ME I'M CONFUSED

just olya [345]

I have no idea I’m doing the same thing rn but good luck bro bro

3 0

1 year ago

Read 2 more answers

the eccentricity of a hyperbola is defined as e=c/a. find an equation with vertices (1,-3) and (-3,-3) and e=5/2

Ksivusya [100]

Answer:

$\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1$ .

Step-by-step explanation:

If (α, β) are the coordinates of the center of the hyperbola, then its equation of the hyperbola is $\frac{(x - \alpha )^{2} }{a^{2} } - \frac{(y - \beta )^{2} }{b^{2} } = 1$ .

Now, the vertices of the hyperbola are given by (α ± a, β) ≡ (1,-3) and (-3,-3)

Hence, β = - 3 and α + a = 1 and α - a = -3

Now, solving those two equations of α and a we get,

2α = - 2, ⇒ α = -1 and

a = 1 - α = 2.

Now, eccentricity of the hyperbola is given by $b^{2} = a^{2}(e^{2} - 1) = 4[(\frac{5}{2} )^{2} -1] = 21$ {Since $e = \frac{5}{2}$ given}

Therefore, the equation of the given hyperbola will be

$\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1$ . (Answer)

6 0

2 years ago

Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).

NemiM [27]

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

$1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}$

Now, differentiate to obtain

$\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}$

$=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}$

To obtain the required form, extract a factor in both the numerator and denominator:

$\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)$

3 0

2 years ago

Copper can be shaped into various forms due to a physical property called

Anni [7]

Answer:

IT CAN BE MANY THINGS

Step-by-step explanation:

5 0

2 years ago