What is the solution set of the equation j+4/j+2=2−1/j ? Note: j≠0, −2

Mathematics

1 answer:

Hunter-Best [27]4 years ago

6 0

Answer:

values of j are: 2 , -1

Step-by-step explanation:

We have been given:

$\frac{j+4}{j+2}={2}-\frac{1}{j}$

We will cross multiply in the above equation we get:

$\frac{j+4}{j+2}=\frac{2j-1}{j}$

$j^2+4j=2j^2+4j-j-2$

$2j^2-j^2+4j-4j-j-2$

$j^2-j-2$

We will solve the quadratic equation by factorizing:

$j^2-2j+j-2$

$j(j-2)+1(j-2)$

$(j+1)(j-2)$

hence, the values of j are: 2 , -1

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In the trapezoid ABCD (AB∥CD) point M∈AD, so that AM:MD=3:5. Line L ∥AB and going through point M intersects diagonal AC and leg

siniylev [52]

Answer:

$\dfrac{AP}{PC}=\dfrac{3}{5}$

$\dfrac{BN}{CN}=\dfrac{3}{5}$

Step-by-step explanation:

Consider triangles AMP and ADC. In these triangles,

angle A is the common angle, so $\angle MAP\cong \angle DAC$ by reflexive property;
angles AMP and ADC are congruent as corresponding angles when two parallel lines MP and CD are cut by transversal AD.

Hence, triangles AMP and ADC are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{AM}{AD}=\dfrac{AP}{AC}\\ \\\dfrac{3x}{3x+5x}=\dfrac{AP}{AC}\\ \\\dfrac{AP}{AC}=\dfrac{3}{8}\Rightarrow AP=\dfrac{3}{8}AC\\ \\PC=AC-AP=AC-\dfrac{3}{8}AC=\dfrac{5}{8}AC,$

$\dfrac{AP}{PC}=\dfrac{\frac{3}{8}AC}{\frac{5}{8}AC}=\dfrac{3}{5}$

Consider triangles ACB and PCN. In these triangles,

angle C is the common angle, so $\angle ACB\cong \angle PCN$ by reflexive property;
angles ABC and PCN are congruent as corresponding angles when two parallel lines PN and AB are cut by transversal BC.

Hence, triangles ACB and PCN are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{CP}{AP}=\dfrac{CN}{CB}\\ \\\dfrac{5x}{3x+5x}=\dfrac{CN}{CB}\\ \\\dfrac{CN}{CB}=\dfrac{5}{8}\Rightarrow CN=\dfrac{5}{8}CB\\ \\BN=BC-CN=BC-\dfrac{5}{8}BC=\dfrac{3}{8}BC,$