All categories

3 years ago

Whats these -4e-9=19 answer

Mathematics

1 answer:

kherson [118]3 years ago

8 0

Its e=7

-4e-9=19
      +9+9
            28
   -4e=28
    -4x7=28-9=19

You might be interested in

Find the general form, Ax2+Ay2+Dx+Ey+F=0, by identifying the coefficients A,D,E,&F center:(0,1); r=1 (x-0)2+(y-1)2=1 x2+y2-2

siniylev [52]

Answer:

The value of A is 1, D is 0, E is -2 and F is 0.

Step-by-step explanation:

The given equation is

$Ax^2+Ay^2+Dx+Ey+F=0$ ...(1)

The standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is the center of the circle and r is the radius.

$(x-0)^2+(y-1)^2=1$

$x^2+y^2-2y+1-1=0$

$x^2+y^2-2y=0$

It can be written as

$x^2+y^2+0x-2y+0=0$ ....(2)

On comparing (1) and (2) we get.

$A=1$

$D=0$

$E=-2$

$F=0$

Therefore the value of A is 1, D is 0, E is -2 and F is 0.

6 0

3 years ago

Read 2 more answers

Why are 0.02 and 0.020 equal?

Marina CMI [18]

The extra 0 doesn’t equal to anything

5 0

3 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$