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tresset_1 [31]
3 years ago
5

The product of two mixed numbers that are each between 5 and 4 is the product between 16 and 25

Mathematics
1 answer:
Alchen [17]3 years ago
4 0
Hmm
that is a true statement, the product of any 2 mixed numbers between 4 and 5, will be between 16 and 25

we can show that it is true by testing the bounds
if the number is 4 and the other is 4, then it is 16
we aren't allowed to do 4 since it isn't mixed and it equals 4
that is the lower bbound

if the numbes 5 and 5 are used, then it is 25
we arene't allowed to use 5 since it isn't mixed and it equals 5

so the product of any 2 numbers between 4 and 5 will be between 16 and 25

so pick any random one

4 and 3/4 times 4 and 5/323
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Identify the values that should be written to complete the X diagram.
cestrela7 [59]

Answer:

See the attachment for what goes on your X diagram.

Rewrite: x² -7x +4x -28

Grouping: (x² -7x) +(4x -28) = x(x -7) +4(x -7) = (x +4)(x -7)

Step-by-step explanation:

The given quadratic is ...

... x² -3x -28 . . . . . a=1, b=-3, c=-28

a) The value at the top of the X diagram is the product a·c = 1·(-28) = -28.

The value at the bottom of the X diagram is the coefficient b = -3.

The values on the sides of the diagram are the factors of -28 that add up to make -3. These are -7 and 4. That is, ...

(-7)·(4) = -28

(-7)+(4) = -3

b) Since the two values on the sides of the diagram add up to give "b", the value of "b" in the equation can be rewritten as the sum of these two numbers. Doing that, we have ...

... x² -3x -28

... = x² +(-7+4)x -28

... = x² -7x +4x -28 . . . . . . order does not matter. It could also be x² +4x -7x -28

c) We can group pairs of terms in the rewritten expression and factor each pair.

... = (x² -7x) +(4x -28) . . . . . first pair has a common factor of x; second pair, 4

... = x(x -7) +4(x -7) . . . . . . . these terms now have a common factor: (x -7)

... = (x +4)(x -7) . . . . . . . . . . the complete factorization of x² -3x -28

3 0
3 years ago
Select the equivalent expression
slava [35]
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7 0
3 years ago
Evaluate the following expressions if a = 2, b = 3, x = 4, and y = 5.<br> b^2+3(2x-y)
butalik [34]

Hi there! Hopefully this helps!

-------------------------------------------------------------------------------------------------------

<h2 /><h2>Answer: 18</h2><h2>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</h2><h2 />

First we need to substitute the value of the variable into the expression.

b^{2}+3(2x-y) = 3^{2} + 3 (2(4)-5) .

Now we need to solve 3^{2} + 3 (2(4)-5).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3^{2} + 3 (2(4)-5)

First, calculate 3 to the power of 2 and get 9.

9+3(2\times 4-5)=18

Multiply 2 and 4 to get 8.

9+3(8-5) =18  

Subtract 5 from 8 to get 3.    

9+3\times 3 =18  

Multiply 3 and 3 to get 9.

9+9 =18.

<h2>And the answer is, you guessed it, 18!</h2>

7 0
3 years ago
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