Which basic geometric figure is labeled with either one capital letter or three capital letters?

Mathematics

1 answer:

tester [92]3 years ago

3 0

Answer-

An angle is labeled with either one capital letter or three capital letters

Solution-

Point-

As a point is 0 dimensional, so you just need 1 point to define a point.

Line-

As a line is 1 dimensional,so you need 2 points to define a line.

Plane-

A plane is 2 0r 3 dimensional, so you need 3 points to define a plane.

Angle-

An angle measures the amount of turn. An angles is labeled with a single letter at the vertex, as long as it is perfectly clear that there is only one angle at this vertex. Otherwise, it is named according to the intersecting lines with 3 letters.

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Which shows the expression below in simplified form?

Kisachek [45]

Answer: C. 3.2 × 10^4

Step-by-step explanation:

for this case the multiplication of the values with exponent are summed, as the exponent are the 6 and -3

6 - 3 = 3

this will be the exponent of the common value which is the 10.

after you only need to multiply 8 and 4 by separate the

8 * 4 = 32.

then we only need to place the values one by one

32 × 10^3

3 is the result of the sum of the exponent.

10 is the common value.

32 is the result of the product.

if we add another zero to the multiplication we get 3.2 and exponent finish with one more zero, then we get this result

3.2 × 10^4

because of that the C is the correct answer, keep in mind that the other answers do not apply correctly the rule of the exponent, because the rest of answer give us wrong answer with final result of the exponent.

3 0

3 years ago

Please I need help with differential equation. Thank you

Inga [223]

1. I suppose the ODE is supposed to be

$\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)$

Solving for $\dfrac{\mathrm dy}{\mathrm dt}$ gives

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}$

which is undefined when $t=\pm1$ . The interval of validity depends on what your initial value is. In this case, it's $t=-\dfrac12$ , so the largest interval on which a solution can exist is $-1\le t\le1$ .