Consider the functions given below

Mathematics

1 answer:

sashaice [31]4 years ago

3 0

I wish I knew the answer but I don’t, but you got this dude but the

ANSWER:y intercept is 4

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Cos(2x) - 3cos(x) + 2= 0 ??

alexandr1967 [171]

Ignore above
your equation is 2(cosx)^2-3cosx+1=0
use quadratic formula to get results
you will end up with
cosx=1 or cosx=1/2
so, x=0 or x=pi/3

3 0

4 years ago

Suppose a normal distribution has a mean of 50 and a standard deviation of 3. What is P(x≤53)?

Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

Use the formula to find the z-score, then go to the table that will give you the probability that the value is less than this z-score.

$z=\frac{53-50}{3}=1$

The probability that the value lies to the left of 1 on a standard bell curve is 84.1345%, or 84%, the third choice down.

5 0

3 years ago

Hey, Can someone help me?

Pepsi [2]

I add up all of the terms that is given and got that Darrel wins.

7 0

3 years ago

In the trapezoid ABCD (AB∥CD) point M∈AD, so that AM:MD=3:5. Line L ∥AB and going through point M intersects diagonal AC and leg

siniylev [52]

Answer:

$\dfrac{AP}{PC}=\dfrac{3}{5}$

$\dfrac{BN}{CN}=\dfrac{3}{5}$

Step-by-step explanation:

Consider triangles AMP and ADC. In these triangles,

angle A is the common angle, so $\angle MAP\cong \angle DAC$ by reflexive property;
angles AMP and ADC are congruent as corresponding angles when two parallel lines MP and CD are cut by transversal AD.

Hence, triangles AMP and ADC are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{AM}{AD}=\dfrac{AP}{AC}\\ \\\dfrac{3x}{3x+5x}=\dfrac{AP}{AC}\\ \\\dfrac{AP}{AC}=\dfrac{3}{8}\Rightarrow AP=\dfrac{3}{8}AC\\ \\PC=AC-AP=AC-\dfrac{3}{8}AC=\dfrac{5}{8}AC,$

$\dfrac{AP}{PC}=\dfrac{\frac{3}{8}AC}{\frac{5}{8}AC}=\dfrac{3}{5}$

Consider triangles ACB and PCN. In these triangles,

angle C is the common angle, so $\angle ACB\cong \angle PCN$ by reflexive property;
angles ABC and PCN are congruent as corresponding angles when two parallel lines PN and AB are cut by transversal BC.

Hence, triangles ACB and PCN are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{CP}{AP}=\dfrac{CN}{CB}\\ \\\dfrac{5x}{3x+5x}=\dfrac{CN}{CB}\\ \\\dfrac{CN}{CB}=\dfrac{5}{8}\Rightarrow CN=\dfrac{5}{8}CB\\ \\BN=BC-CN=BC-\dfrac{5}{8}BC=\dfrac{3}{8}BC,$