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Art [367]
3 years ago
12

Explain why the polynomial 100x^2 + 150x + 49 is not a perfect square.

Mathematics
1 answer:
Ad libitum [116K]3 years ago
4 0
Answer: The correct answer is Choice C.

For this polynomial to be a perfect square, it would need to be:
(10x + 7)^2

This will ensure that the first terms and the last terms will be 100x^ and 49. However, if you use foil to multiply the factors, you will not get 150x for the center term. Choice C also states that 150x will not be the middle term.
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Help asap show work
katovenus [111]

Answer:  see below

<u>Step-by-step explanation:</u>

17) 5n - 3 > -2n + 4     and    9 - 4n ≥ -2n + 1

   <u> +2n   </u>    <u> +2n      </u>              <u>    +2n </u>   <u>+2n      </u>

   7n - 3   >     4                    9 - 2n  ≥       1

        +3         +3                  -9                 -9

    7n       >      7                       -2n   ≥     -8

  <u> ÷7     </u>    <u>     ÷7  </u>                   <u> ÷ -2 </u>  ↓  <u> ÷ -2 </u>

          n  >     1         and            n   ≤   4                  

Graph:   o----------------- ·

              1                     4

18) 9k - 2 ≤ 9 + 10k < 9 - 4k

     9k - 2 ≤ 9 + 10k      and     9 + 10k < 9 - 4k

  <u> -10k     </u>    <u>     -10k </u>               <u>       +4k </u>   <u>    +4k  </u>

     -k - 2 ≤ 9                             9 + 14k < 9

   <u>      +2</u>   <u>+2         </u>                 <u> -9          </u>  <u>-9         </u>

     -k      ≤  11                                  14k  <  0

 <u> ÷ -1    </u>   ↓<u> ÷ -1  </u>                      <u>       ÷14 </u>    <u>÷14      </u>

          k  ≥ -11                and               k < 0

Graph:    · -----------------o

              -11                  0    

19) 4 - n ≤ 10 + 5n ≤ 6n + 1

     4 - n ≤ 10 + 5n       and     10 + 5n ≤  6n + 1

   <u>    -5n </u>   <u>      -5n  </u>                 <u>      -6n </u>   <u>-6n     </u>

   4 - 6n ≤ 10                           10  -  n  ≤       1    

  <u>-4        </u>  <u> -4            </u>              <u> -10        </u>    <u>  -10  </u>

        -6n ≤ 6                                    -n ≤   -9

   <u>  ÷ -6   </u>↓ <u>÷ -6     </u>                    <u>    ÷ -1 </u>↓<u> ÷ -1    </u>

           n ≥ -1              and                 n ≥ 9

Graph:    · -----------------→

              -9                  

20) 4p - 5 ≤ 2p + 9 ≤ 4p + 7

     4p - 5 ≤ 2p + 9     and     2p + 9 ≤ 4p + 7

  <u> -2p       </u>   <u>-2p    </u>                 <u> -4p    </u>   <u>-4p      </u>

    2p - 5 ≤         9                -2p + 9  ≤        7    

    <u>      +5   </u>  <u>    +5    </u>             <u>        -9  </u>    <u>     -9  </u>

       2p  ≤       14                   -2p       ≤      -2

   <u>  ÷ 2   </u>   <u>    ÷ 2     </u>               <u>÷ -2    </u> ↓<u>    ÷ -2    </u>

         p ≤       7          and         p    ≥      1

Graph:    · ----------------- ·

              1                    7

6 0
3 years ago
Please help!!!!
DochEvi [55]

Answer:

256h28k8

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
The list shows the score of each game completed at a bowling alley during a one-hour period. 90, 96, 120, 124, 130, 135, 140, 14
HACTEHA [7]
90, 96, 120, 124, 130, 135, 140, 145, 148, 290, 290. How would the data graphed on a horizontal number line appear to be clustered or skewed?
6 0
2 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
What is the equation to this?
Fiesta28 [93]
THE EQUATION TO THIS PROBLEM IS y=2x+12
7 0
3 years ago
Read 2 more answers
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