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3 years ago

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A prime polynomial is an irreducible polynomial with integer coefficients that cannot be factored into polynomials of lower degr

ee over the real number system. Which trinomial is prime?
A.) x2 – 7x – 18
B.) x2 – 9x – 18
C.)x2 + 3x – 18
D.)x2 + 17x – 18

2 answers:

ELEN [110]3 years ago

8 0

your answer is

b. x²-9x-18

have a good day :) :) :)

SpyIntel [72]3 years ago

4 0

The answer is B.) x^2 - 9x - 18.

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Help me right now its easy 30pts

kotykmax [81]

Answer:

See the attachment

Step-by-step explanation:

7. The common difference is 4, so the next term will be 17+4 = 21, and the term after that will be 21+4 = 25. Figuring the next two terms as 21, 25 is sufficient to select the appropriate answer.

8. The table entry (0, 14) tells you only choices G and J are viable. The difference between terms of 22-14 = 8 tells you the appropriate choice is the one with a slope of 8, choice G.

9. In a linear equation, the sum of exponents of the variables in any/every term must not be greater than 1. (All exponents must be integers, as they are here.) Selections A and D have squared terms, so are not linear. Selection C has the product of variables "xy", so is not linear.

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3 years ago

Suppose that 13 inches of wire costs 78 cents.

GarryVolchara [31]

Answer:

$0.42

Step-by-step explanation:

create a proportion:

13/.78 = 7/x

cross-multiply to get:

13x = 7(.78)

13x = 5.46

x = 5.46/13

x = .42

5 0

2 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

<u>Zero Product Property</u>: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the <u>Zero Product Property</u>, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$

Therefore, the solutions to the given equation are: x = 1 or x = -1

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1 year ago

What is the measure of the following ∠? m∠AOD? 40 80 120

stepladder [879]

120 is the answer im pretty sure

5 0

3 years ago

Read 2 more answers

What is the scale factor

nadezda [96]

Answer:

2/5

Step-by-step explanation:

Since it's from a big square to a smaller square, the smaller number will be at the top. Also, use the same side to find the scale factor, but since this is a square and all sides of a square are equal, it doesn't matter but in other shapes it does so practice i guess:

Left side of big square: 5

Left side of small square: 2

So scale is 2/5

To confirm:

5 · 2/5 = 10/5 = 2, we're right

5 0

3 years ago

Read 2 more answers