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3 years ago

Write a trinomial expression in standard form that has a degree of 5

1 answer:

3 0

$\text{We need to find a trinomial expression in standard form with degree 5.}\\
\\
\text{we know that a trinomial expression is an expression which consists}\\
\text{three terms in it.}\\
\\
\text{also given that the degree of the trinomial is 5, so the highest exponent}\\
\text{appear in the trinomial would be 5.}\\$

$\text{so if we write a trinomial expression in standard form with degree 5,}\\
\text{the first term must be of the form }x^5.\\
\\
\text{next two terms can be any smaller exponent term or any constant.}\\
\\
\text{thus one such trinomial expression with degree 5 is: }\\
\\
x^5-3x^2+2$

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We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance

MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

$h(t) = -16\cdot t^{2}+256\cdot t+1680$ (1)

Where:

$h$ - Height above the ground, measured in feet.

$t$ - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

$h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680$

$h(3\,s) = 2304\,ft$

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at $t = 5\,s$ and $t = 7\,s$ :

$h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680$

$h(5\,s) = 2560\,s$

$h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680$

$h(7\,s) = 2688\,s$

We notice that the small object ascends in the given interval.

$\Delta h = h(7\,s)-h(5\,s)$

$\Delta h = 128\,ft$

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that $h = 2640\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t-960=0$ (2)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 10\,s$ and $t_{2}= 6\,s$

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that $h = 0\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t +1680 = 0$ (3)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 21\,s$ and $t_{2} = -5\,s$

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0

3 years ago

3^(x−1) = 27 I need the answer please and thanks

bogdanovich [222]

Make 27 into 3^3, then since the bases are the same, solve (x-1) = 3. The answer is 4.

6 0

4 years ago

Read 2 more answers

Suppose that you draw two cards from a deck. After drawing the first card, you do not put the first card back in the deck. What

S_A_V [24]

Answer:

(B) 0.0588

Step-by-step explanation:

The probability is calculated as a division between the number of possibilities that satisfy a condition and the number of total possibilities. Then, the probability that the first card is diamonds is:

$P_1=\frac{13}{52}$