Plz help fast <br> I will mark Brainlyist

What is product of 3x-4 an 5x^2-2x+6 <br> help me please, please, please

Fofino [41]

(3x-4)(5x^2-2x+6)
15x^3-6x^2+18x-20x^2+8x-24
15x^3-26x^2+26x-24

4 0

3 years ago

Sarah and Jenna are paid by the hour at their jobs. Their earnings, in dollars, for different numbers of hours worked are shown

Ivan

Answer:

1

Step-by-step explanation:

6 0

3 years ago

Two cones are cut by a plane as shown, where the plane is perpendicular to the bases. The cross-section of the cones and the pla

Ira Lisetskai [31]

Answer:

D) a hyperbola

Step-by-step explanation:

IMO, the image is not helpful, as it suggests the ends of the curve are nearly parallel to each other. In fact, the ends of the curve asymptotically approach straight lines parallel to the sides of the cone. That is, there are asymptotes that form an X shape. This is a characteristic of a hyperbola.

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A parabola is formed when the plane cuts the cone parallel to one side. That cut plane only intersects one nap of the cone. At a shallower angle yet, one gets an ellipse. Parallel to a base, one gets a circle.

3 0

3 years ago

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

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

so

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

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

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