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Anna35 [415]
3 years ago
13

When is it most appropriate to use the completing the square method to solve a quadratic equation?

Mathematics
1 answer:
mash [69]3 years ago
4 0

Answer:

Don't forget to include a ± sign in your equation once you have taken the square root. Next, if the coefficient of the squared term is 1 and the coefficient of the linear (middle) term is even, completing the square is a good method to use. Finally, the quadratic formula will work on any quadratic equation.

Step-by-step explanation:

x² + 4x = 15 can be solved best using the completing the square method.

2x(x−3)=0 can be best solved using the zero product property.

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Write -41.27 as a mixed number.
Salsk061 [2.6K]

Answer:

-41 27/100

Step-by-step explanation:

4 0
3 years ago
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
A man that is 6 feet tall is standing so that the tip of his shadow is 20 feet from a light pole. His shadow is 8 feet long. Wha
pychu [463]

Answer:

The answer to your question is: the height of the light pole is 15 ft.

Step-by-step explanation:

See the picture below

Now, we can do proportions to finds the height of the light pole

                   \frac{20}{8} =  \frac{x}{6}

Solve for x

                      x = 6(20) / 8

                     x = 120 / 8

                     x = 15 ft

                   

4 0
3 years ago
4[5(12+3)-2]-7 please help I'm stuck on this fifth grade math problem
ehidna [41]
<em>two hundred and eighty five</em>
3 0
3 years ago
Read 2 more answers
Somebody help me with this please .
erastova [34]

Step-by-step explanation:

12. Cos 60° = 8/c

0,5 = 8/c

0,5 c = 8

c = 16

D² = V16²-8²

= V256-64

=V192 = V16×12 = 4V12

= 4V4×3 = 8V3

13. Cos 30° = 6/b

V3/2 = 6/b

V3 b = 12

b = 12/V3

b/Sin B = a /sin A

b/Sin90° = 6/ sin 60°

<u>b</u> = <u> </u><u> </u><u> </u><u>6</u><u> </u><u> </u><u> </u>

1 V3/2

b× <u>V3</u> = 6

2

b = 6× 2/V3

= 12/V3

8 0
4 years ago
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