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andre [41]
3 years ago
15

Which of the following is a factor of x2 − 21x + 110?

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
8 0
X² - 21x + 110
first you break it up into two parenthesis & since x is squared you need two x's 
since 110 is positive and 21 is negitive, both of your signs need to be negitive
(x - ?)(x - ?)
then you need to think! what are the factors of 110?
Which of these factors add up to 21? those factors go in the parenthesis

Does this Help? Let me know if you need anything!
shtirl [24]3 years ago
4 0

Answer:

Option B is correct. (x-10)

Step-by-step explanation:

The given term is = x^{2} -21x+110

Now, to know the factors of this term, we will split 110 in a way that the factors sum upto 21.

So, 110 can be split as 11\times10 and the sum of 11 and 10 is 21.

Now, we will split the whole term as ;

x^{2} -10x-11x+110=0

x(x-10)-11(x-10)=0

(x-11)(x-10)

So, the factors of x^{2} -21x+110 are : (x-11) and (x-10)

Therefore, out of the given options, option B is correct.

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Gemiola [76]

Answer:

B

Step-by-step explanation:

Prime Factorization involves breaking down a number into prime numbers (aka numbers that can only be divided by itself and 1)

36

9  x  4

(3x3) x (2x2)

Hope that helps!

3 0
3 years ago
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25x -5 + 32 = (when x = 3)
Ugo [173]

Answer:

102

Step-by-step explanation:

25 x 3 - 5 + 32

75 - 5 + 32

70 + 32

=102

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3 years ago
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In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
3 years ago
Sean places 28 tomato plants in rows. All rows contain the same number of plants. There are netween 5 and 12 plants in each row.
IrinaVladis [17]

Answer:

4 rows of 7 plants each

Step-by-step explanation:

To find the number of plants in each row between 5 and 12 requires us to find the factors of the total 28.

28 has factors 1,28 and 2,14 and 4,7. Only 4,7 has a number between 5 and 12. There are 4 rows of 7 plants.

7 0
3 years ago
hey, can someone please double check these questions for me / or help me solve them if they are incorrect? thank you U-U ​
zzz [600]

Question 11

The directrix is a horizontal line, which means the parabola opens either upward or downward. In this case, it opens downward. This is because all answer choices have a negative leading coefficient. Also, it's because the focus is below the directrix.

For vertically opening parabolas, we use this form

4p(y-k) = (x-h)^2

where (h,k) is the vertex and p is the focal distance, aka the distance from the vertex the focus. To find (h,k), we start at the focus (0,-4) and move directly up until we reach the directrix y = 4. We'll arrive at (0,4). The midpoint of (0,-4) and (0,4) is (0,0) which is the vertex's location. So (h,k) = (0,0).

Note that in moving from (0,-4) to (0,4) is a span of 4 units. So this is the value of p.

Plug h = 0, k = 0, p = 4 into the equation mentioned and solve for y

4p(y-k) = (x-h)^2

4*4(y-0) = (x-0)^2

16y = x^2

y = (1/16)x^2

The only adjustment we need to make is to change the 1/16 to -1/16 so that the parabola opens downward.

<h3>Answer:  Choice D.  y = -(1/16)x^2</h3>

===============================================

Question 3

The given equation is in the form y = ax^2+bx+c

In this case,

  • a = 2
  • b = 4
  • c = 3

Let's compute the x coordinate of the vertex h

h = -b/(2a)

h = -4/(2*2)

h = -1

This h value is plugged into the original function to find k

f(x) = 2x^2+4x+3

f(-1) = 2(-1)^2+4(-1)+3

f(-1) = 1

We find that h = -1 and k = 1 pair up together. In short, (h,k) = (-1,1) is the vertex.

<h3>Answer: Choice B.  (-1,1)</h3>
8 0
2 years ago
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