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3 years ago

What expression is equivalent to 3-(-2+5)

Mathematics

1 answer:

maxonik [38]3 years ago

4 0

Answer:

Ok ok, I see. So the answer is -9.

Step-by-step explanation:

We know the expresion is:

-3(-2+5)

We need to distrubute the -3 to what is in the parthese.

When I say distribute, basically you need to multiply the numbers inside the parethese by -3. So we can think of it as:

-3*-2=6

and:

-3*5=-15

Which we can rewrite as:

(6-15)

Subtract 15 from 6 and you get -9:

6-15=-9

Sorry hope this helps!

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137x8 multiply with expanded form please help 10 ponte for your help please help!

Step2247 [10]

Answer: maybe 20

Step-by-step explanation:

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3 years ago

What is the complete factorization of p(x)=32x5y−2xy5 over the integers?

egoroff_w [7]

Answer:

$p(x)=2xy(2x-y)(2x+y)(4x^2+y^2)$

Step-by-step explanation:

$p(x)=32x^5y-2xy^5=2xy(16x^4-y^4)=2xy(4x^2-y^2)(4x^2+y^2)\\\\\boxed{p(x)=2xy(2x-y)(2x+y)(4x^2+y^2)}$

The factoring of the difference of squares is applicable:

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3 years ago

Find the distance between the two points. (1, 2), (7, 6) The distance between the two points is __ units

katrin [286]

Answer:

7.2

Step-by-step explanation:

distance between the two points. (1, 2), (7, 6) = √(x-x')²+(y-y')²

d = √(7-1)²+(6-2)² = √52 = 7.2

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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

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

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

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

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