Ask question

Ask question

All categories

2 years ago

11

What is the coefficient of the second term of the polynomial? (In other words, what is B?

1 answer:

Scrat [10]2 years ago

7 0

Answer: The coefficient of the second term of the polynomial is B=5

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

You might be interested in

Which similarity postulate or theorem can be used to verify that the two

Elanso [62]

SAS

ASA

SSS

AAA

These are similarity criterion

3 0

3 years ago

Read 2 more answers

Help me with this question please ? and how do I get the answer?

melamori03 [73]

1) you have to find the difference between pipe A and pipe B, which is 6.40.

23.55-17.15=6.40

2) after finding the difference. You will take that difference and put it out of 23.55 and multiply by 100 to get your percentage. As so:

(6.40/23.55)(100)= 27.176% which rounds to 27% as a whole number.

Your answer is 27%. Hope that helps!

7 0

3 years ago

This is flipping a figure over the axis of symmrtry

fgiga [73]

Answer:

Step-by-step explanation:

FSDGGHJKL

4 0

3 years ago

More math help pls. :)

Step2247 [10]

By using trigonometric relations, we will see that x = 9.97°.

<h3>How to find the missing angle?</h3>

First, we need to find the bottom cathetus of the smaller triangle, we will use the relation:

Tan(θ) = (opposite cathetus)/(adjacent cathetus).

Where:

θ = 26°
Adjacent cathetus = k
Opposite cathetus = 55ft.

Replacing that we get:

Tan(26°) = 50ft/k

Solving this for k, we get:

k = 55ft/tan(26°) = 112.8 ft

Now, we can see that the longer triangle adds 200ft to this cathetus, so now we will have:

angle = x
opposite cathetus = 55ft
adjacent cathetus = 112.8ft + 200ft = 312.8ft.

Then we have:

Tan(x) = (55ft/312.8ft)

Using the inverse tangent function in both sides, we get:

x = Atan(55ft/312.8ft) = 9.97°

If you want to learn more about right triangles, you can read:

brainly.com/question/2217700

6 0

2 years ago

2x^3-x^2-3x=210<br> the answer is 5 but I want to know why.

mel-nik [20]

Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

-------------------------------------------------------------------------

$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

-----------------------------------------------------------------------

$9x^{2}$ $-3x$ $-210$

--------------------------------------------------------------------------

$42x$ $-210$

$42x$ $-210$

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

4 0

2 years ago

Read 2 more answers