3 years ago

What is the coefficient of the term of degree 5 in the polynomial below

Mathematics

2 answers:

Olegator [25]3 years ago

5 0

I think the answer is a)4.

ioda3 years ago

4 0

Answer is 4, because its the number directly in front of x^5

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Simplify -3p 3 + 5p + (-2p 2) + (-4) - 12p + 5 - (-8p 3).

MissTica

Answer:5p^3-2p^2-7p+1

Step-by-step explanation:

-3p^3+5p+(-2p^2)+(-4)-12p+5-(-8p^3)

open brackets

-3p^3+5p-2p^2-4-12p+5+8p^3

Collect like terms

-3p^3+8p^3-2p^2+5p-12p-4+5

5p^3-2p^2-7p+1

3 0

3 years ago

Find x from given equations sin30 + 2 cot^2 30 + x cos^2 30= 8+ tan^2 45 + cos 60

julia-pushkina [17]

Answer

1/2+2×(sqrt3)^2+x.(sqrt3/2)^2=8+1^2+1/2

1/2+2×3+x.3/4=8+1+1/2

1/2-1/2+6+x.3/4=9

x.3/4=9-6

x.3×4=3

x.3=3×4

x.3=12

x=12/3

x=4.

so, the value of x is 4.

hope it helps you.

4 0

3 years ago

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3 years ago

The mall is 4.72 kilometers from Julie’s house and 1.83 kilometers from Taylor’s house. How much farther does Julie live from th

blondinia [14]

Julies lives 2.89 kilometers more from the mall than Taylor.

7 0

3 years ago

Read 2 more answers

Find the third side in simplest radical form:<br> 25

Gre4nikov [31]

Answer:

<h3> $\boxed{ \bold{24}}$ </h3>

Step-by-step explanation:

$\mathsf{given}$

$\mathsf{hypotenuse(h) = 25}$

$\sf{perpendicular (p) = 7}$

$\sf{base(b) = }$ ?

Now, Using Pythagoras theorem

$\sf{{h}^{2} = {p}^{2} + {b}^{2} }$

plug the values

⇒ $\sf{ {25}^{2} = {7}^{2} + {b}^{2} }$

Evaluate the power

⇒ $\sf{625 = 49 + {b}^{2} }$

Swap the sides of the equation

⇒ $\sf{49 + {b}^{2} = 625}$

Move constant to right hand side and change it's sign

⇒ $\sf{ {b}^{2} = 625 - 49}$

Calculate the difference

⇒ $\sf{ {b}^{2} = 576}$

Squaring on both sides

⇒ $\sf{b = 24}$

Hope I helped!

Best regards!

8 0

3 years ago