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1 year ago

Which expression is equivalent to (3x4+17x3−x)+(−2x4−5x)?

2 answers:

7 0

Answer
X=-6/55
Step-by-step explanation:
(3×4+17×3-x)+(-2×4-5x)?
=12+51-x-8-5x
=12+51-x-8-5x
=63-x-8-5x
=63-8-x-5x
=55-x-5x
=55-6x
X=-6/55
I hope its help u... Thanks

umka2103 [35]1 year ago

5 0

The equivalent expression of the given expression (3x⁴ + 17x³ − x) and (− 2x⁴ − 5x) will be (x⁴ + 17x³ − 6x).

<h3>What is an equivalent function?</h3>

The equivalent is the functions that are in different forms but are equal to the same value.

The expression is given below.

(3x⁴ + 17x³ − x) + (− 2x⁴ − 5x)

Then the equivalent expression of the given expression will be

3x⁴ + 17x³ − x − 2x⁴ − 5x

x⁴ + 17x³ − 6x

The equivalent expression of the given expression (3x⁴ + 17x³ − x) and (− 2x⁴ − 5x) will be (x⁴ + 17x³ − 6x).

More about the equivalent link is given below.

brainly.com/question/889935

#SPJ2

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A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co

butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

$\displaystyle \vec{v}=\frac{\vec{x}}{t}$

Where $\vec x$ is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

$\displaystyle \vec{x}=\vec{v}.t$

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_1}=\ km/h$

The displacement of the plane in 2 hours is

$\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)$

$\displaystyle \vec{x_1}=km$

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

$\displaystyle \vec{v_2}=$

$\displaystyle \vec{v_2}=km/h$

The displacement in 1 hour is

$\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)$

$\displaystyle \vec{x_2}=km$

The total displacement is the vector sum of both

$\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+$

$\displaystyle \vec{x_t}=km$

$\displaystyle \vec{x_t}=$

The distance from the airport is the module of the displacement:

$\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}$

$\displaystyle |\vec{x_t}|=894.43\ km$

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3 years ago

On Saturday Stella worked for two hours helping her mother in the house. she spent 1/3 of her time doing laundry, 1/4 of a time

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$\bf \stackrel{\textit{total hours time}}{2}-\cfrac{1}{3}-\cfrac{1}{4}\impliedby \textit{we'll use an LCD of \underline{12}}
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2-\cfrac{1}{3}-\cfrac{1}{4}\implies \cfrac{(12\cdot 2)~-~(4\cdot 1)~-~(3\cdot 1)}{12}\implies \cfrac{24-4-3}{12}\implies \cfrac{17}{12}
\\\\\\
\textit{12 goes into 17 \underline{1 time} and it has a \underline{remainder} of 5 }\implies 1\frac{5}{12}$

which is one hour and 25 minutes.

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3 years ago

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Answer:

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