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valkas [14]
3 years ago
6

Please help, i really dont understand this. will give brainliest. There are 6 dogs and 5 cats.

Mathematics
1 answer:
pav-90 [236]3 years ago
4 0

Answer:

13) 11C5 or 11C6 = 462

14) it appears to be only 1

DCDCDCDCDCD

15) 9 choose 6 or 9 choose  3 = 84

Step-by-step explanation:

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How do you solve 2/3x+15=17? it is confusing. i have been stuck on this problem for a while...
kirill [66]
Find the opposite of adding 15. So it would be subtracting 15. So subtract 15 on both sides. So it would have 2/3x on the left and then 2 on the other side. (2/3x=2) Now find the opposite of multiplying 2/3. It is dividing 23/3. So divide 23/3 on both sides. So multiply 2/3 times 1/2. It would equal 3. x = 3
Hope this helps. :-)
3 0
3 years ago
Find the equation of the line in slope-intercept form containing the points (6, -1) and (-3, 2).
egoroff_w [7]
Slope intercepf is y=mx+b wherem=slope and b= y intercept
slope is found by doing
(y1-y2)/(x1-x2)
points are (6,-1) and (-3,2)
(x,y)
x1=6
y1=-1
x2=-3
y2=2
subsitute
(-1-2)/(6-(-2))=-3/(6+2)=-3/8
slope=-3/8
subsitute
y=-3/8x+b
subsitute and solve for b
(-3,2)
x=-3
y=2
2=-3/8(-3)+b
2=9/8+b
2=16/8
subtract 9/8 from both sides
16/8-9/8=b
7/8=b
y=-3/9x+7/8 is the equation
5 0
3 years ago
Marissa bought 2 and a half gallons of orange juice to make punch how many quarts of orange juice did. Marissa buy
Nookie1986 [14]
The answer is as follows:
There are 2 pints in a quart. 
There are 2 quarts in a half gallon.
There are 2 half gallons in a gallon.
Therefore, we can conclude that there are 10 quarts of orange juice.
7 0
2 years ago
Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}
miskamm [114]

Answer:

-3+i\sqrt{3} , 1+\sqrt{3}

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

\alpha = x+iy\\\beta = x-iy

since they are conjugates

\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}

\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}

Imaginary part of the above =0

i.e. \sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1

So the value of alpha = -3+i\sqrt{3} , 1+\sqrt{3}

3 0
3 years ago
PLEASE HELP THIS IS A QUESTION OF LIFE OR DEATH
den301095 [7]
C because the x-axis and the y-axis are switched, basically.
4 0
3 years ago
Read 2 more answers
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