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Marat540 [252]
3 years ago
10

2 1/3 ÷y=1 2/5 ÷(−1.1)

Mathematics
1 answer:
Arada [10]3 years ago
7 0

Answer:

1.2x - 10.8 - 2.3x - 9.2 = -1.1x - 20; Rewrite as follows. (x 2y)(xy 2) = (x 2 x)(y y 2) ... x + 2.3; - x / 5 = 2; (x - 4) / (- 6) = 3; (-3x + 1) / (x - 2) = -3; x / 5 + (x - 1) / 3 = 1/5 ... 2 x 2 - 8 = 0; x 2 = -5; 2x 2 + 5x - 7 = 0; (x - 2)(x + 3) = 0; (x + 7)(x - 1) = 9; x(x - 6) = -9 ... For what value of the constant k does the quadratic equation x 2 +2x = - 2k .

Step-by-step explanation:

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Express the vector negative 3 Bold i plus 6 Bold j plus 2 Bold k as a product of its length and direction.
ANTONII [103]

Answer:

A

Step-by-step explanation:

6 0
3 years ago
Aubrey went shopping at her favorite store target she bought three shirts and a pair of jeans that cost $28 she spent a total of
aliya0001 [1]

Answer:

Each shirt cost $20

Step-by-step explanation:

88- $28= 60

60/3= $20

8 0
2 years ago
A+1=1 find the value of b^a​
Ray Of Light [21]

Answer:

1

Step-by-step explanation:

a+1=1

a=1-1

a=0

b^a which means:

b^0

whenever power is 0 answer is always one

so value of b^a is 1

6 0
3 years ago
Read 2 more answers
Subtract <br><br> \frac{3n+2}{n-4}-\frac{n-6}{n+4}
aalyn [17]
(3n+2)/(n-4) - (n-6)/(n+4)

common denominator (n-4)(n+4)

{(n+4)(3n+2)-(n-4)(n-6)}/{(n-4)(n+4)}

Use the foil method:

{(3n²+14n+8)-(n²-10n+24)}/{(n-4)(n+4)}

distribute negative sign:

{(3n²+14n+8-n²+10n-24)}/{(n-4)(n+4)}

subtract:

(2n²+24n-16)/{(n-4)(n+4)}

take out 2:

2{n²+12n-8}/{(n-4)(n+4)}

 
4 0
3 years ago
Read 2 more answers
In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In ad
zysi [14]

Answer:  a) 0.0792   b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)   (1)

Similarly ,

P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

5 0
3 years ago
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