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Softa [21]
3 years ago
8

I apologize for all the questions I asked ​

Mathematics
1 answer:
Paul [167]3 years ago
3 0

Answer:

Step-by-step explanation:

It's not clear but yeah it's fine if you ask a lot cause everytime I see this I know it's you :)

You might be interested in
The function f(x) = (x - 4)(x - 2) is shown.
agasfer [191]

Answer:

all real numbers greater than or equal to -1

Step-by-step explanation:

In order to solve this problem we need to know the vertex and the direction its pointing.

First we expand,

x^2-6x+8

To find the x value of the vertex we use this formula (-b/2a).

-(-6)/2(1) = 3

Now we plug 3 in the equation to get the y value,

(3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1

The vertex is (3,-1)

We know the graph points up because x^2 is positive

The vertex is the lowest point, so we now know that -1 is the starting range and if the graph is pointing up, that means all values greater than -1.

This leads to our answer all real numbers greater than or equal to -1.

6 0
3 years ago
Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp
iren [92.7K]
I’m pretty sure it’s C and E
4 0
3 years ago
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
3 years ago
Jelly ate 3 apples. This represents 50% of the bag of apples. How many apples were originally in the bag
rjkz [21]
<h3>Answer:  6 apples</h3>

x = number of apples originally

50% of x = 0.50*x = amount of apples eaten = 3

0.50x = 3

x = 3/0.50 ....... divide both sides by 0.50

x = 6

You can think of 50% as 1/2. If 3 apples represent half the bag, then the full bag is 6 apples. So half of 6 is 3.

6 0
3 years ago
Read 2 more answers
Can you plz solve this ​
Lesechka [4]

Answer:

a) acute angle

b) obtuse angle

c) right angle

d) reflex angle

e) straight line angle

f) acute angle

Step-by-step explanation:

Acute angle

  • angle that is smaller than 90°
  • 0° < θ < 90°

Right angle

  • angle that is 90°
  • shaped of a "L"

Obtuse angle

  • angle that is greater than 90° but smaller than 180°
  • 90° < θ < 180°

Straight line angle

  • angle that is 180°
  • drawn in a straight line

Reflex angle

  • angle than is greater than 180°
  • θ > 180°
8 0
3 years ago
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