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djyliett [7]
3 years ago
5

Using the greatest common factor for the terms, how can you write 56 + 32 as a product?

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0
56 + 32 = (8 x 7) + (8 x 4) = 8(7 + 4) (Answer C)


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Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet
frosja888 [35]

There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy

• x^3 divides f(x)

• (x-1)^3 divides f(x)^2

I'll also assume <em>f(x)</em> is monic, meaning the coefficient of the leading term is 1, or

f(x) = x^5 + \cdots

Since x^3 divides f(x), and

f(x) = x^3 p(x)

where p(x) is degree-2, and we can write it as

f(x)=x^3 (x^2+ax+b)

Now, we have

f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2

so if (x - 1)^3 divides f(x)^2, then p(x) is degree-2, so p(x)^2 is degree-4, and we can write

p(x)^2 = (x-1)^3 q(x)

where q(x) is degree-1.

Expanding the left side gives

p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2

and dividing by (x-1)^3 leaves no remainder. If we actually compute the quotient, we wind up with

\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}

If the remainder is supposed to be zero, then

\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}

Adding these equations together and grouping terms, we get

(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1

Then b=-1-a, and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1

So, we end up with

p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}

6 0
2 years ago
A plane can fly 540 miles with the wind in one hour less than it can fly 480 miles against the wind. The average wind speed is 3
Bess [88]

Answer:

The speed of the plane in still air = 150 mph

Step-by-step explanation:

This is a relative velocity question

Let the velocity of the plane in still air be v

And let the time the plane can fly 480 miles against the wind be t

(Velocity of the plane relative to the wind) = (velocity of plane) - (velocity of wind)

Flying against the wind

(Velocity of plane relative to the wind) = (480/t)

(Velocity of the plane) = v

(Velocity of the wind) = 30 mph

(480/t) = v - 30

t = 480/(v-30) (eqn 1)

Flying with the wind

(Velocity of plane relative to the wind) = 540/(t-1)

(Velocity of the plane) = v

(Velocity of the wind) = -30 mph

540/(t - 1) = v + 30

t - 1 = 540/(v+30) (eqn 2)

Since t is equal in both cases, substitute the value of t in eqn 1 into eqn 2.

[480/(v-30)] - 1 = [540/(v+30)]

Multiply through by (v+30)(v-30)

480(v+30) - [(v+30)(v-30)] = 540(v-30)

480v + 14400 - (v² - 900) = 540v - 16200

480v + 14400 - v² + 900 = 540v - 16200

v² + 540v - 480v - 16200 - 14400 - 900 = 0

v² + 60v - 31500 = 0

Solving the quadratic equation,

v = 150 mph or v = -210 mph

We'll pick the positive answer because of the directions we have established.

Therefore, the speed of the plane in still air = 150 mph

Hope this Helps!!!

3 0
2 years ago
Hikers spent the following amounts of time (in minutes) to complete a nature hike:48, 46, 52, 57, 58, 52, 61, 56.
Solnce55 [7]
For A for the mean just add them all up which would be which is 430 and divide by the number of numbers there is so 430 divided by 8 = 53.75 and the median order them from least to greatest and find the one in the middle which is 52 and 56 when there's two just add them and then divide it by 2 which is 54

Hope this helped :)
8 0
3 years ago
Read 2 more answers
How to do question 22?
Aleks04 [339]

Answer:

A = 20sinθ(6 + 5 cosθ)  cm²

Step-by-step explanation:

Drop perpendiculars DE and CF to AB.

Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.

The formula for the area of the trapezium is

A = ½(a + b)h

DE = 10sinθ

AE = 10cosθ

BF = 10cosθ

EF = CD = 12 cm

AB = AE + EF  + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ

A = ½(a + b)h

   = ½(12 +12 + 20 cosθ) × 10 sinθ

   =(24 + 20 cosθ) × 5 sinθ

   = 4(6 + 5cosθ) × 5sinθ

   = 20sinθ(6 + 5 cosθ)  cm²

6 0
3 years ago
Can someone help I have to be done by tomorrow morning plz order them from least to greatest plz
Sindrei [870]
11. 0.03, 3/10,1 1/8,1 5/6,2
7 0
3 years ago
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