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3 years ago

Can i get the answer quick

Mathematics

1 answer:

Kruka [31]3 years ago

6 0

Answer:

p^3 - 7p^2q^2 + 2pq

Step-by-step explanation:

3pq +5p^2q^2 + p^3 = p^3 + 5p^2q^2 + 3pq

Let the number be x.

x + p^3 + 5p^2q^2 + 3pq + p^3 - pq

= 3p^3 - 2p^2q^2 + 4pq

x + p^3 + p^3 + 5p^2q^2 + 3pq - pq

= 3p^3 - 2p^2q^2 + 4pq

x + 2p^3 + 5p^2a^2 + 2pq

= 3p^3 - 2p^2q^2 + 4pq

= 3p^3 - 2p^2q^2 + 4pq - (2p^3 + 5p^2q^2 + 2pq)

= 3p^3 - 2p^2q^2 + 4pq - 2p^3 - 5p^2q^2 - 2pq

= 3p^3 - 2p^3 - 2p^2q^2 - 5p^2q^2 + 4pq - 2pq

= p^3 - 7p^2q^2 + 2pq

Therefore,

p^3 - 7p^2q^2 + 2pq should be added.

You might be interested in

Look at the right-angled triangle ABC.

olga_2 [115]

Answer:

∠x = 90°

∠y = 58°

∠z = 32°

Step-by-step explanation:

he dimensions of the angles given are;

∠B = 32°

Whereby ΔABC is a right-angled triangle, and the square fits at angle A, we have;

∠A = 90°

∠B + ∠C = 90° which gives

32° + ∠C = 90°

∠C = 58°

∠x + Interior angle of the square = 180° (Sum of angles on a straight line)

∠x + 90° = 180°

∠x = 90°

∠x + ∠y + 32° = 180° (Sum of angles in a triangle)

90° + ∠y + 32° = 180°

∠y = 180 - 90° - 32° = 58°

∠y + ∠z + Interior angle of the square = 180° (Sum of angles on a straight line)

58° + ∠z +90° = 180°

∴ ∠z = 32°

∠x = 90°

∠y = 58°

∠z = 32°

8 0

3 years ago

1. Find the measure of angle p.

geniusboy [140]

Answer:

Step-by-step explanation:

To find exterior angles you add non-adjacent interior angles.

8 0

3 years ago

(PRECALC) I need help on this one plz ❤️❤️❤️ help a girl out & the favor will be returned

tia_tia [17]

Answer:

The coordinates of the P' point are (-17.3,-10).

Step-by-step explanation:

We can use the components of the vector from 0 to P point. We know that the diameter is 40 meters, so the magnitude of the vector will be 20 m.

Now, the angle from the negative x-direction to the vector is 210°-180° = 30°.

The x-component of the vector will be:

$V_{x}=-Vcos(30)$

$V_{x}=-20cos(30)=-17.32\: m$

It is negative because the component is in the negative x-direction.

The y-component of the vector will be:

$V_{y}=-Vsin(30)$

$V_{y}=-20sin(30)=-10\: m$

It is negative too because the component is in the negative y-direction.

Therefore, the coordinates of the P' point are P'(-17.3,-10)

I hope it helps you!

7 0

3 years ago

Find the value of <br>sin 300°.cos 390°-cot 150°.sin 120°

aleksandr82 [10.1K]

note that :

cot150 = 1/tan150

cos390 = cos(360+30) = cos30

6 0

4 years ago

Read 2 more answers

PLEASEEEE HELPPPPPPPPPPP

andreyandreev [35.5K]

Answer:

The correct option is first graph.

Therefore,

The two points for the line y+5=-2(x-4) ( red color )line are

point A( x₁ , y₁) ≡ ( 0 , 3) .......Blue color

point B( x₂ , y₂) ≡ $(\dfrac{3}{2},0)$ ............Green color

The Graph is attached below.

Step-by-step explanation:

Given:

$y+5=-2(x-4)$

Which can also be written as

$y=-2x+8-5\\y=-2x+3$ .....Equation of line

Let the points be point A, and point B

To Find:

point A( x₁ , y₁) ≡ ?

point B( x₂ , y₂) ≡ ?

Solution:

For Drawing a graph we require minimum two points but we will have here three points.

For point A( x₁ , y₁)

Put x = 0 in the given equation we get

y = -2 × 0 +3

y =3

∴ y = 3

∴ point A( x₁ , y₁) ≡ ( 0 , 3)

For point B( x₂ , y₂)

Put y = 0 in the given equation we get

0 = -2x + 3

$x=\dfrac{3}{2}$

∴ point B( x₂ , y₂) ≡ $(\dfrac{3}{2},0)$

Therefore,

The two points for the line y+5=-2(x-4) ( red color )line are

point A( x₁ , y₁) ≡ ( 0 , 3) .......Blue color

point B( x₂ , y₂) ≡ $(\dfrac{3}{2},0)$ ............Green color

The Graph is attached below.

6 0

4 years ago