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muminat
3 years ago
7

How do you add 5x/2x-16 + 2/x^2-64

Mathematics
2 answers:
BabaBlast [244]3 years ago
8 0

Answer:

5x^4−160x^2+4/2x^2

Step-by-step explanation:

5x/2  x−16+2/x^2−64


5/2x^4−80x^2+2/x^2


5x^4−160x^2+4/2xx


5x^4−160x^2+4/2x^2




zvonat [6]3 years ago
4 0
Try using the app Socratic, I love it soooooo muchhhhhh
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the pie chart shows the information about the favourite ice cream flavour of each student in a school there are 720 students in
kirill115 [55]

Answer:

chocolate= 180 students

vanilla= 120

strawberry= 210

mango= 210

Step-by-step explanation:

The chocolate section of the pie chart is at a right angle (90 degrees), which means a quarter of the students prefer chocolate

720/4= 180

The vanilla section is 60 degrees which is 2/3 of 90

180/3=60

60x2=120

the mango and strawberry sections represent whats left which is 420 students. The sections are equal to each other so they're each 210 students

7 0
3 years ago
(x + 4)2 – 3(x + 4) – 3 = 0 Select the solution(s) of the original equation.
Aneli [31]

Answer:

x=−7

Step-by-step explanation:

(x+4)(2)−3(x+4)−3=0

(x)(2)+(4)(2)+(−3)(x)+(−3)(4)+−3=0(Distribute)

2x+8+−3x+−12+−3=0

(2x+−3x)+(8+−12+−3)=0(Combine Like Terms)

−x+−7=0

4 0
2 years ago
Plz Help me due today??!!! I dont under stand???
Triss [41]

Answer: Median = 44

lower quartile = 38

upper quartile= 55

Step-by-step explanation:

3 0
3 years ago
Hi please please help me
Kaylis [27]

Answer:

141.12 cm^2

Step-by-step explanation:

   Since the area of a trapezoid is the average of the bases multiplied by the height, we can just plug these numbers into the equation! Since the average of 10.2 and 12.2 is 11.2, and the height is 12.6, we multiply them to get the area to be 141.12 cm^2.

6 0
3 years ago
Read 2 more answers
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
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