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kirill115 [55]
3 years ago
15

Solve 10-9x^2+4x=-6x^2

Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
6 0

Answer

i found it on the internet

babymother [125]3 years ago
3 0
The answer is x = (2 +/- √34)/3
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Which graph represents y=1/2x+2 <br><br> please help!!
wel

Answer:

B

Step-by-step explanation:

Let’s first look at the y-intercept. Since the equation is in y = mx + b form, we know that m is the slope and b is the y-intercept, so the slope is 1/2 and the y-intercept is 2, or (0,2). We can narrow down our search by noticing that C and D don’t intersect (0,2). That leaves A and B.

The slope is calculated by (y_1 - y_2)/(x_1 - x_2) given points (x_1, y_1) and (x_2, y_2). The slope for A is (4-0)/(2-(-2)) = 4/4 = 1 and the slope for B is (2-0)/(0-(-4)) = 2/4 = 1/2. The slope of B matches the slope that we are looking for. So, the answer is B.

8 0
3 years ago
Is 10.2 x 10 written in scientific notation. EXPLAIN
Virty [35]

Answer:

No

Step-by-step explanation:

Scientific notation has to be a number between 1 and 10 times 10 to the something power. 10.2 is not between 1 and 10

7 0
2 years ago
Read 2 more answers
5m - 4n, when m=5 and n = 3.
tiny-mole [99]

Answer:

5m - 4n

5(5) - 4(3) ( As m= 5 and n= 3)

25 - 12

= 13

8 0
2 years ago
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What do you add 15/4 to make 4
koban [17]
The answer would be 1/4 or \frac{1}{4} i hope this answer helps!
8 0
3 years ago
Please Help ASAP!!
Brilliant_brown [7]

Answer:

The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.

Step-by-step explanation:

In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't.  Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype.  However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.  

4 0
3 years ago
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