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3 years ago

For the polynomial below,what is the power of the term with the coefficient of 1? X^3+1/3x^4+6x+5

Mathematics

2 answers:

julia-pushkina [17]3 years ago

8 0

Answer:

It's 3

Step-by-step explanation:

valentina_108 [34]3 years ago

5 0

Original Polynomial
${x}^{3} + \frac{1}{ {3x}^{4} } + 6x + 5$
Coefficient Concept Expansion
$1 \times {x}^{3} = {x}^{3}$
The Power of the term with 1 as the coefficient is 3.

A coefficient is a number multiplied by some variable. For instance, 4x has a coefficient of 4 or 56x has a coefficient of 56.

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3x + 11 - 2y + 4x - 6 + 5y <br><br> please solve!!! just simplify by combining like terms

rusak2 [61]

the answer to your question is
7y+7x+5

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4 years ago

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What is the sign of the product (3)(-3)(-2)(4)? (5 points) Select one: a. Positive, because the products (3)(-3) and (-2)(4) are

Gnesinka [82]

Answer: a. Positive, because the products (3)(-3) and (-2)(4) are negative, and the product of two negative numbers is positive

Step-by-step explanation: It is also possible to rearrange the factors.

(3)(4) = 12 a positive number.

(-3)(-2) = 6, a positive number. Again "a negative times a negative is a positive" --Jaime Escalante, <em>Stand and Deliver</em>

Now you have two positive numbers: 12×6 = 72

The product is positive either way.

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3 years ago

Use the property of exponents to rewrite the expression<br> (-4qr)(-4qr)(-4qr)(-4qr)

ivanzaharov [21]

Answer:

(-4qr)^4

Step-by-step explanation:

(-4qr)(-4qr)(-4qr)(-4qr)

There are 4 sets of -4qr being multiplied together

(-4qr)^4

5 0

3 years ago

Help? with complete solution please help me please

serious [3.7K]

9514 1404 393

Answer:

13 ft
(a) 1 second; (b) t = 0, t = 1/2

Step-by-step explanation:

Let w represent the length of the wire. Then the height of attachment is (w-1). The Pythagorean theorem tells us a relevant relation is ...

5² +(w -1)² = w²

w² -2w +26 = w² . . . . . . . eliminate parentheses, collect terms

26 = 2w . . . . . . . . . . . . add 2w

13 = w . . . . . . . . . . . . divide by 2

The length of the wire is 13 feet.

(a) When h = 0, the equation is ...

0 = -16t^2 +8t +8

Dividing by -8 puts this into standard form:

2t^2 -t -1 = 0

Factoring, we get ...

(2t +1)(t -1) = 0

The positive value of t that makes a factor zero is t = 1.

It will take 1 second for the gymnast to reach the ground.

(b) When h = 8, the equation is ...

8 = -16t^2 +8t +8

Subtract 8 and divide by 8 to get ...

0 = -2t^2 +t

0 = t(1 -2t) . . . . factor out t

Values of t that make the factors zero are ...

t = 0

t = 1/2

The gymnast will be 8 feet above the ground at the start of the dismount, and 1/2 second later.

8 0

2 years ago

D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

4 0

3 years ago