X=2 y=5 2x^0y^-2 Please help

Use the discriminant to determine how many real number solutions exist for the quadratic equations -4j+3j-28=0

aleksandr82 [10.1K]

the correct equation is
- 4j² + 3j - 28 =0

ax² + bx + c,
a = -4, b=3 c =- 28.
The discriminent D = b² - 4.a.cD = 3² - 4(-4)(-28)D = -439
If D ≥ 0 there are either 2 roots or one double if = 0)if D< 0, there are no real roots (but 2 imaginary roots)Since D < 0, then there is no real roots
The answer there is no real roots

8 0

3 years ago

6th grade math help me, please:)))

Lesechka [4]

Step-by-step explanation:

Hi there!!! 

The answer is option A. 

reason look in picture. 

if you want to solve it just cross multiply it,

 $60x = 240000 \times 100$ 

or,

 $x = 24000000 \div 60$ 

and the answer would be 400000.

Hope it helps...

5 0

3 years ago

Assume that different groups of couples use the XSORT method of gender selection and each couple gives birth to one baby. The XS

olganol [36]

Answer:

Mean and Standard deviation for the numbers of girls in groups of 36 births are 18 and 3 respectively.

Step-by-step explanation:

We are given that he X SORT method is designed to increase the likelihood that a baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5.

Now consider a group consisting of 36 couples.

The above situation can be represented through binomial distribution;

$P(X=r)=\binom{n}{r} \times p^{r} \times (1-p)^{n-r} ;x=0,1,2,3,.....$

where, n = number of trials (samples) taken = 36 couples

r = number of success

p = probability of success which in our question is probability

of a girl, i.e.; p = 0.5

Let X = Numbers of girls in groups of 36 births

So, X ~ Binom(n = 36, p = 0.5)

Now, mean for the numbers of girls in groups of 36 births is given by;

Mean, E(X) = $n \times p$ = $36 \times 0.5$ = 18

Also, standard deviation for the numbers of girls in groups of 36 births is given by;

Standard deviation, S.D.(X) = $\sqrt{n\times p\times (1-p)}$

= $\sqrt{36\times 0.5\times (1-0.5)}$

= 3

3 0

2 years ago

Diagram shows a section of a bridge between the points A and K. The length of line segment is 640 meters. ∆ABC, ∆CDF, and ∆FJK a

Marta_Voda [28]

AK = 640
Δ ABC and ΔFJK are similar. They are small triangles.
ΔCDF is the big triangle.

640 / 2 = 320 m= CF
AC & FK= 320/2 = 160 m each

2AC = CF = 2FK
2(160) = 320 = 2(160)

BG = 20 m

20/160 = x / 320
20*320 = 160x
6400 = 160x
6400/160 = x
40 = x

Area of CDF = (40 m * 320 m)/2 = 12,800 / 2= 6,400 m²

3 0

2 years ago

1.

Anna11 [10]

Answer:

y/x

Step-by-step explanation:

If x is proportional to y, then:

y = kx, where k is a constant

This can be rearranged to give:

k = y/x

As mentioned, k is a constant, therefore, the answer is y/x

7 0

1 year ago