BRAINLIST Julius needs to determine how much information needs to be given about an isosceles triangle to be able to draw an isosceles triangle. He is looking at six different cases and is trying to determine if that amount of detail will allow him to construct just one isosceles triangle or if it allows for more than one isosceles triangle.

Case 1. The length of the two congruent sides and the measure of the included angle.

Case 2. The lengths of two non-congruent sides and the measure of the included angle.

Case 3. The lengths of two non-congruent sides.

Case 4. The measure of the two congruent angles and the included side.

Case 5. The measure of two non-congruent angles.

Case 6. The measure of two non-congruent angles and the included side.

Place each case in under the correct heading in the table.

Forms only one isosceles triangle
Forms more than just one isosceles triangle

step-by-step explanation:

what grade level is this? need to know so i can look in my old textbooks!

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: