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ankoles [38]
3 years ago
15

Find x and y (Please answerrr) ​

Mathematics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer: idont even know

Step-by-step explanation:

rfcgggggggggggggggggggdrthyjcgfdtey

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Write an equation in point-slope form of the line that passes through the two given points. Use the first point to write the equ
a_sh-v [17]
Answer:
(-1,6)
Explanation:
use (y1-y2), (x1-x2)
5 0
2 years ago
The average weight of the class of 35 students was 45 KG. with the admission of a new student the average weight came down to 44
melamori03 [73]

The weight of the new student is 27 kg.

Average weight

= total weight ÷total number of students

<h3>1) Define variables</h3>

Let the total weight of the 35 students be y kg and the weight of the new student be x kg.

<h3>2) Find the total weight of the 35 students</h3>

<u>45 =  \frac{y}{35}</u>

y= 35(45)

y= 1575 kg

<h3>3) Write an expression for average weight of students after the addition of the new student</h3>

New total number of students

= 35 +1

= 36

Total weight

= total weight of 35 students +weight of new students

= y +x

44.5  =  \frac{y + x}{36}

<h3>4) Substitute the value of y</h3>

44.5 =  \frac{1575 + x}{36}

<h3>5) Solve for x</h3>

36(44.5)= 1575 +x

1602= x +1575

<em>Subtract 1575 from both sides:</em>

x= 1602 -1575

x= 27

Thus, the weight of the new student is 27 kg.

7 0
2 years ago
For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec
telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

4sin(x^2) = 0

sin(x^2) = 0

x^2 = n\pi where n = 0, 1, 2,3, 4, 5,...

x = \sqrt(n\pi)

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0

xcos(x^2) = 0

x = 0 or

cos(x^2) = 0

x^2 = \frac{\pi}{2} + n\pi where n = 0, 1, 2, 3

x = \sqrt{\pi(n+1/2)}

Since we are restricting x between 0 and 3 we can stop at n =  2

So our critical points are at

x = 1.25, y = f(1.25) = 4sin(1.25^2) = 4

x = 2.17 , y = f(2.17) = 4sin(2.17^2) = -4

x = 2.8, y = f(2.8) = 4sin(2.8^2) = 4

For the inflection point, we can take the 2nd derivative and set it to 0

f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0

cos(x^2) = 2x^2sin(x^2)

tan(x^2) = \frac{1}{2x^2}

We can solve this numerically to get the inflection points are at

x = 0.81, y = f(0.81) = 4sin(0.81^2) = 2.44

x = 1.81, y = f(1.81) = 4sin(1.81^2) = -0.54

x = 2.52, y = f(2.52) = 4sin(2.52^2) = 0.27

3 0
3 years ago
Given the function f(x) in red, write the equation for the transformed function g(x)in green
STatiana [176]
g(x)=(x-4)^3-5

3 0
1 year ago
-7(n - 2) + 2n = -5(n + 6)
skad [1K]

Given the expression below

-7(n-2)+2n=-5(n+6)

To find n

Open the brackets

\begin{gathered} -7(n-2)+2n=-5(n+6) \\ -7n+14+2n=-5n-30 \\ -7n+2n+14=-5n-30 \\ -5n+14=-5n-30 \end{gathered}

Collect like terms

\begin{gathered} -5n+14=-5n-30 \\ -5n-(-5n)=-30-14 \\ 0\ne-44 \end{gathered}

Since, the sides are not equal,

Hence, there is no solution

7 0
1 year ago
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