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AnnZ [28]
3 years ago
7

Find two consecutive integers whose sum is 67.

Mathematics
1 answer:
Vlada [557]3 years ago
3 0
Let the smaller integer be x.
The next larger consecutive integer is 1 more than x, so it's x + 1.

x + x + 1 = 67

2x + 1 = 67

Answer is option B.
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If x+1 and x-1 are the factors of the polynomial ax^3 + x^2 - 2x +b,find the values of a and b
Nadusha1986 [10]

The polynomial remainder theorem says that dividing a polynomial p(x) by x-c leaves a remainder of p(c)=0 if x-c is a factor of p(x). In this case, check c=-1 and c=1.

a(-1)^3+(-1)^2-2(-1)+b=-a+b+3=0

a(1)^3+(1)^2-2(1)+b=a+b-1=0

From the first equation,

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and substituting into the second gives

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3 0
3 years ago
Could someone please help me?
elena-14-01-66 [18.8K]

Answer:

(-6,0),(-5,-1),(-1,-2)

6 0
3 years ago
6x + 5y = 147<br> 8x + 4y = 156
satela [25.4K]

Answer:

[12, 15]

Step-by-step explanation:

{6x + 5y = 147

{8x + 4y = 156

-¾[8x + 4y = 156]

{6x + 5y = 147

{-6x - 3y = -117 ← New Equation

________________

2y = 30

___ ___

2 2

y = 15 [Plug this back into both equations to get the x-coordinate of 12]; 12 = x

So, using the Elimination Method, what I did here was took the bottom equation and multiplied it by -¾, turning 8x to -6x, canceling out all <em>x-terms</em><em>.</em><em> </em>It does not matter which variable you choose to eliminate, as long as you know what you are doing.

I am joyous to assist you anytime.

3 0
3 years ago
Please help! factorise 3e^2 + 5e
Oksana_A [137]

3e^2 + 5e = e(3e +5)

............................

3 0
3 years ago
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