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Tpy6a [65]
3 years ago
11

What are four consecutive odd integers with the sum of 232

Mathematics
1 answer:
zvonat [6]3 years ago
7 0
Four consecutive odd integers, eh? Let x be the first of these odd integers. The next odd integer is thus <span>x+2</span>, and the next is <span>x+4</span>, and the last is <span>x+6</span>. The sum of all four of these is 232, so you have <span><span><span>x+x+2+x+4+x+6=232</span><span>4x+12=232</span></span></span>
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B is the midpoint of AC. AB = 2x+12 and BC = 5x + 10. Find AC
3241004551 [841]

Answer:

x=6

Step-by-step explanation:

/AB/=2x+12

/BC/=5x+10

3x+2=5x-10 subtract 2 from both sides

3x=5x-12 subtract 5x from both sides

-2x=-12 divide both sides by -2x

x =6

5 0
2 years ago
9. Consider the function f(x) = x² + 3x – 4.
BaLLatris [955]

Answer:

  c.  f(x) = (x + 4)(x - 1)

Step-by-step explanation:

Since you're familiar with the product of two binomials:

  (x +a)(x +b) = x² + (a+b)x + ab

you know that the constants in the binomial factors must ...

  • have a product of -4
  • have a sum of +3

__

All of the choices except B have binomial constants that have a product of -4.

In order, the sums of the remaining choices are ...

  A: 1-4 = -3

  C: 4-1 = 3 . . . . this is the correct choice

  D: -2+2 = 0

6 0
3 years ago
X4 – 17x2 + 16 = 0 Rewrite the equation in terms of u
slavikrds [6]
You probably are interested in expressing the given equation as a quadratic equation in u, as it will make it easy to find the solutions.

Let u = x²
So,
u² = x⁴

So, the given equation can be written as:

u² - 17u + 16 = 0

Now the equation is quadratic in u and the solutions can be calculated using quadratic formula or factorization.

6 0
3 years ago
Read 2 more answers
(URGENT) need help with this so badly someone help asap
Fynjy0 [20]
The answer for A is 2,026. I set a porportion up which is 77/x = 3.8/100 

The answer to B is 26

8 0
3 years ago
Find the number of integral solutions of x+y +z = 12, where −3 ≤ x ≤ 4, 2 ≤ y ≤ 11, and<br> z ≥ 3.
SashulF [63]

There are 59 integer solutions

Such questions are best solved by writing cases and calculating the total number of cases. So beginning with

1)  x = -3. The possible combinations are as follows:-

-3 2 13

-3 3 12

-3 4 11

-3 5 10

-3 6 9

-3 7 8

-3 8 7

-3 9 6

-3 10 5

-3 11 4

10 combinations

2) x = -2

-2 2 12

through

-2 11 3

10 combinations

3) x = -1

-1 2 11

through

-1 10 3

9 combinations

4) x = 0

0 2 10

through

0 9 3

8 combinations

as we can see from the pattern at x =1  we get 7 combinations, at x =2  we get 6 combinations, at x=3 we get 5 combinations and at x =4  we get 5 combinations.

Thus total number of combinations 4+5+6+7+8+9+10+10 = 59 integer solution.

Learn more about combinations here :

brainly.com/question/13387529

#SPJ1

5 0
2 years ago
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