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zaharov [31]
3 years ago
11

Half of the product of two consecutive numbers is 105. To solve for n, the smaller of the two numbers, which equation can be use

d?
A.) n^2 + n – 210 = 0
B.) n^2 + n – 105 = 0
C.) 2n^2 + 2n + 210 = 0
D.) 2n^2 + 2n + 105 = 0
Mathematics
2 answers:
solniwko [45]3 years ago
7 0

Answer:

n^2+n-210=0

Step-by-step explanation:

alexdok [17]3 years ago
6 0
There are several information's already given in the question. Based on those information's the answer can be detrmined.
Smaller of the two integers = n
Larger of the two integers = n + 1
Then
(1/2) * [(n) * ( n + 1)] = 105
(1/2) * (n^2 + n) = 105
m^2 + n = 210
n^2 + n - 210 = 0
From the above deduction, it can be concluded that the correct option among all the options that are given in the question is the first option or option "A".
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1st. (2 root 2 + root 3 ) ( 2 root 3 - root 2)
satela [25.4K]

Answer:

1st: 3*root6 + 5

2nd: 35*root2 + 115

3rd: 24*root2 - 20*root6 + 15*root3 - 18

4th: 17*root6 - 38

5th: 13*root10 - 42

Step-by-step explanation:

To simplify these expressions we need to use the distributive property:

(a + b) * (c + d) = ac + ad + bc + bd

So simplifying each expression, we have:

1st.

(2 root 2 + root 3 ) ( 2 root 3 - root 2)

= 4*root6 - 2*2 + 2*3 - root6

= 3*root6 - 4 + 9

= 3*root6 + 5

2nd.

(root 5 + 2 root 10) (3 root 5 + root 10)

= 3 * 5 + root50 + 6*root50 + 2*10

= 15 + 5*root2 + 30*root2 + 100

= 35*root2 + 115

3rd.

(4 root 6 - 3 root 3) (2 root 3 - 5)

= 8*root18 - 20*root6 - 6*3 + 15root3

= 24*root2 - 20*root6 + 15*root3 - 18

4rd.

(6 root 3 - 5 root 2 ) (2 root 2 - root 3)

= 12*root6 - 6*3 - 10*2 + 5*root6

= 17*root6 - 18 - 20

= 17*root6 - 38

5th.

(root 10 - 3 ) ( 4 - 3 root 10)

= 4*root10 - 3*10 - 12 + 9*root10

= 13*root10 - 30 - 12

= 13*root10 - 42

8 0
3 years ago
PLEASE HELP PRECALCULUS <br> SEE ATTACHMENT
koban [17]

Answer:

1+6i

Step-by-step explanation:

Given:

f(x)=x^4 - 2x^3 + 38x^2 - 2 + 37

zero of f(x) = 1-6i

another zero of function = ?

Conjugate Zero theorem:

As per conjugate zero theorem, if a function f(x) has real coefficients and one of zero is a complex number then the conjugate of that complex number will also be a zero of that function i.e. complex zeroes will occur in complex conjugate pairs.

conjugate of 1-6i is 1+6i

hence another zero of f(x) will be 1+6i !

3 0
3 years ago
Helppppppppppppppppppppppppppppp Fasttttt
sertanlavr [38]
What you have to do is multiply each of the sides so you have 80, 60, and 48.

Add those up to get 188. Then multiply this by 2 and you get 376.

Your answer is D. (376)

P.S if this helped you out, please give me brainliest if you don't mind. I am only a few more from leveling up :D
5 0
4 years ago
What is the equation of the line that is parallel to the line 5x + 2Y=12 and passesthrough the point (-2,4) ?
sladkih [1.3K]

Answer:

y = -\frac{5}{2}x - 1

Step-by-step explanation:

5x + 2y = 12

2y = -5x + 12

y = -\frac{5}{2}x + 6

y = mx + b

4 = -\frac{5}{2} ( -2 ) + b

4 = 5 + b

b = -1

y = -\frac{5}{2}x - 1

5 0
3 years ago
Joel wants to buy a new tablet computer fram a store having a 20% off sale on all tablets. the tablet he wants has an original c
dusya [7]

Answer:

joel subtract the 5% sales tax.

The correct expression is

1.05(190)(0.8) = $159.60


6 0
3 years ago
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