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abruzzese [7]
3 years ago
8

Numbers 6,7,8 I can't figure out

Mathematics
1 answer:
marysya [2.9K]3 years ago
7 0
6. Take your compass and place the pointed edge on point B. Place one point on each side of B, each the same distance away from B. Next, place your compass on one of the two new points and extend your compass to draw a circle. Repeat with the SAME radian from the other point. Find where the two circles intersect with each other and draw a line from the points of intersection to point B. Place point A anywhere on that line that you just created and then you're done!

7. Select any place along either line and place point S on it. Next, using the same method as above, draw two circles with the same radius around both points S and R. Draw a line through the intersection points. Locate the intersection where your new line connects with the line across from the shared line of RS. Place a point at the intersection, for your reference, then connect that point to point S. Now you have completed this problem as well.

8. Use a straight edge to draw one line. Place points A and B on each end. Use the circle method yet again to find a line perpendicular to line AB. Next, take your compass and set it to the distance from point A to B. Use that same distance to make a point on the perpendicular line. This creates point C. The final step is to connect A with C and B with C.
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dlinn [17]
Picture?? Or I can’t help you
5 0
3 years ago
an appliance store decreases the price of a 19-in. television set 26% to a sale of $463.98. what was the original price?
pogonyaev

100%-26%=74%

x=the original price

\frac{0.74}{1}=\frac{463.98}{x}

\frac{463.98}{0.74}=\frac{0.74x}{0.74}

627=x

The original price was $627.

6 0
3 years ago
Which of the following systems is possible? a system of two linear equations with exactly two solutions a system of two linear e
Basile [38]

Answer:

A system of two linear equations that is independent and has no solution

Step-by-step explanation:

That’s the answer for E2020

4 0
4 years ago
Read 2 more answers
Given: AE ≅ CE ; DE ≅ BE Prove: ABCD is a parallelogram. Parallelogram A B C D is shown. Diagonals are drawn from point A to poi
nikklg [1K]

Answer:

The proof is below

Step-by-step explanation:

Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.

In ΔACD and ΔBEC

AD=BC              (∵Opposite sides of a parallelogram are equal)

∠DAC=∠BCE       (∵Alternate angles)

∠ADC=∠CBE        (∵Alternate angles)

By ASA rule, ΔACD≅ΔBEC

By CPCT(Corresponding Parts of Congruent triangles)

AE=EC and DE=EB

Hence, AE is conruent to CE and BE is congruent to DE

6 0
3 years ago
The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find
9966 [12]

Answer:

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<em> General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given Differential equation  y'' − 5 y' + 4 y = x

Given equation in operator form

        D²y - 5 Dy +  4 y = x

⇒     ( D² - 5 D +  4 ) y =x

⇒    f(D) y = Q

where  f(D) = D² - 5 D +  4 and Q(x) = x

<em>The auxiliary equation  f(m) =0</em>

<em>           m²-5 m + 4 =0</em>

         m² - 4 m - m + 4 =0

        m ( m -4 ) -1 ( m-4) =0

         (m - 1) =0   and ( m-4) =0

        <em> m = 1 and m =4</em>

<em>The complementary function </em>

<em></em>Y_{c} = C_{1} e^{x} + C_{2} e^{4x}<em></em>

<u><em>Step(ii)</em></u>:-

<u><em>particular integral</em></u>

<em>Particular integral</em>

<em>     </em>y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2}  - 5 D +  4} X<em></em>

<em>taking common '4' </em>

<em>                          </em>= \frac{1}{4(1 +  (\frac{D^{2}  - 5 D}{4} ))} X<em></em>

<em>                         </em>

<em>                           </em>=\frac{1}{4}  (1 + (\frac{D^{2} -5D}{4})^{-1} )} X<em></em>

<em>applying binomial expression</em>

<em>      ( 1 + x )⁻¹    = 1 - x + x² - x³ +.....       </em>

<em>                          </em>=\frac{1}{4}  (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X<em></em>

<em>Now simplifying and we will use notation D = </em>\frac{dy}{dx}<em></em>

<em>                        </em>=\frac{1}{4}  (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}<em></em>

<em>Higher degree terms are neglected</em>

<em>                     </em>=\frac{1}{4}  (x - (\frac{ -5 D}{4}) x)<em></em>

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<u><em>Final answer</em></u><em>:-</em>

<em>          General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

<em></em>

<em>         </em>

<em> </em>

     

4 0
3 years ago
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