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Deffense [45]
3 years ago
13

Help me answer and explain STEP BY STEP for #10 and #11

Mathematics
2 answers:
Reptile [31]3 years ago
4 0

these questions don't need steps

madam [21]3 years ago
4 0

Answer:

10. C, Quadratic trinomial

11. D, (6d - 1) (2d + 1)

Step-by-step explanation:

10. 7x² - 5x + 9 is a quadratic trinomial as there are 3 terms in the expression (trinomial)

11. to factor 12d² + 4d - 1 we can use the quadratic formula, factoring, or completing the square. when the first term is not equal to 1, it may be easier to use the quadratic formula or factoring by grouping. in this example, i will use factoring by grouping which is another way to factor. I will write the middle term (4d) as a subtraction using 2 numbers that when multiplied together give us 12. these 2 numbers are 6 and 2, as 6 x 2 = 12 and 6 - 2= 4

we have the following:

12d² + 6d - 2d - 1 < group 2 factors together

(12d² + 6d) (-2d - 1) < solve for each factor by factoring

to factor 12d² + 6d, we need to find a common factor between the 2 numbers. both have the number 6 in common and the letter d, so we can factor out 6d from the expression

6d(2d + 1)

now we will factor (-2d - 1). we want to get 2d + 1 as a result of the factorization we do, so we need to find a number that gives us that. both -2d and -1 have a -1 in common, so we can factor out a -1 from the expression

-1(2d + 1)

we now have the following:

6d(2d + 1) - 1(2d + 1) < since we have 2 (2d + 1), we only need to write one. we can combine 6d and -1 into one factor

(6d - 1) (2d + 1) is our answer. the answer choice would be D

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Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

Third year    (\frac{5}{4} )^{2}x-------------(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x

Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

7 0
3 years ago
Consider a parallelogram with vertices at (0, 3), (2, 0), (4, 2), and (2, 5).
goldenfox [79]

Answer:

B

Step-by-step explanation:

So a reflection over the x is a change in the y value.

Only point (2,0) is on the x-axis, so it will not move.

4 0
2 years ago
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Shkiper50 [21]
If we multiply the bottom equation by 2 and move x to the right, it becomes:

4y = 2x-38

Now we can substitute it for the 4y in the top equation:

3x + (2x-38) = -23 => 5x = -23+38 => 5x = 15 => x=3

Then 4y = 2*3-38 => y = -8

So the solution is (3,-8)
3 0
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Mila [183]

Answer:

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6 0
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Step-by-step explanation:

6 0
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