What is the average(mean)of 7,4,6,8,5

Answers

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. Out of 140 students, 50 passed in English and 20 passed in both Nepali and English. The number of students who passed in Nepali is twice the number of students who passed in English. Using a Venn-diagram, find the number of students who passed in Nepali only and who didn't pass in both subjects.

Answers

Answer:

80 ;

Step-by-step explanation:

Given :

Total number of students = μ = 140

Let :

Number of students who passed in English = E

Number of students who passed in Nepali = N

n(NnE) = 20

n(E) only = n(E) - n(NnE) = 50 - 20 = 30

Students who passed English only = 30

Number of students who passed in Nepali is twice the number who passed in English

n(N) = 2 * n(E) = 2 * 50 = 100

Number of students who passed in Nepali only

n(N) only = n(N) - n(NnE) = 100 - 20 = 80

Students who passed Nepali only = 80

The number who didn't pass both subjects :

μ - (English only + Nepali only + English and Nepali)

140 - (30 + 80 + 20)

140 - 130

= 10

Express sin 404 in terms of the sine of a positive acute angle.

Answers

sin404 = sin(360 + 44)
as sin(360 + α) = sinα
= sin44

How do i simplify -3x-15?

Answers

Answer:

$45$

Step-by-step explanation:

$-3 * -15$

Cancel the negative signs on both sides.

$3 * 15$

Evaluate.

$=45$

Answer: -x-5

Step-by-step explanation:

-3x-15 . Both 3 and 15 are variables of 3, this means they can both be divided by 3. -3/3= -1 and -15/3=-5. So -1x-5, or -x-5

Hope this helped! :)

Simplifying rational expressions1. (c+8)(c-8)/(c-8)(c+3)
2. n^2+4n-12/n^2+2n-8
3. 42x^2y^3/28x^3y
4. m^2-3m-10/m-5

Answers

1. $((c + 8)(c - 8))/((c - 8)(c + 3))_,_$
   $(c + 8)/(c + 3)_,_$

2. $(n^(2) + 4n - 12)/(n^(2) + 2n - 8)_,_$
   $(n^(2) + 6n - 2n - 12)/(n^(2) + 4n - 2n - 8)_,_$
   $(n(n) + n(6) - 2(n) - 2(6))/(n(n) + n(4) - 2(n) - 2(4))_,_$
   $(n(n + 6) - 2(n + 6))/(n(n + 4) - 2(n + 4))_,_$
   $((n - 2)(n + 6))/((n - 2)(n + 4))_,_$
   $(n + 6)/(n + 4)_,_$

3. $(42x^(2)y^(3))/(28x^(3)y)_,_$
   $(3y^(2))/(2x)_,_$

4. $(m^(2) - 3m - 10)/(m - 5)_,_$
   $(m^(2) - 5m + 2m - 10)/(m - 5)_,_$
   $(m(m) - m(5) + 2(m) - 2(5))/(m - 5)_,_$
   $(m(m - 5) + 2(m - 5))/(m - 5)_,_$
   $((m + 2)(m - 5))/(m - 5)_,_$
   $m + 2_,_$

Hugh has to earn at least $400 to meet his fundraising goal. He has only 100 books that he plans to sell at $8 each. Which inequality shows the number of books, x, Hugh can sell to meet his goal?

Answers

If you would like to find the inequality that shows the number of books Hugh can sell to meet his goal, you can do this using the following steps:

x = ? ... the number of books
$8 * x >= $400
8 * x >= 400
x >= 400 / 8
x >= 50 books

The correct result would be: $8 * x >= $400.

Answer:

50 ≤ x ≤ 100

Step-by-step explanation:

The graph of which function has an axis of symmetry at x =-1/4 ?f(x) = 2x2 + x – 1

f(x) = 2x2 – x + 1

f(x) = x2 + 2x – 1

f(x) = x2 – 2x + 1

Answers

we know that

The equation of the vertical parabola in vertex form is equal to

$y=a(x-h)^(2)+k$

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

$x=h$ ------> axis of symmetry of a vertical parabola

we will determine in each case the axis of symmetry to determine the solution

case A) $f(x)=2x^(2)+x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=2x^(2)+x$

Factor the leading coefficient

$f(x)+1=2(x^(2)+0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+0.125=2(x^(2)+0.5x+0.0625)$

$f(x)+1.125=2(x^(2)+0.5x+0.0625)$

Rewrite as perfect squares

$f(x)+1.125=2(x+0.25)^(2)$

$f(x)=2(x+0.25)^(2)-1.125$

the vertex is the point $(-0.25,-1.125)$

the axis of symmetry is

$x=-0.25=-(1)/(4)$

therefore

the function $f(x)=2x^(2)+x-1$ has an axis of symmetry at $x=-(1)/(4)$

case B) $f(x)=2x^(2)-x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=2x^(2)-x$

Factor the leading coefficient

$f(x)-1=2(x^(2)-0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+0.125=2(x^(2)-0.5x+0.0625)$

$f(x)-0.875=2(x^(2)-0.5x+0.0625)$

Rewrite as perfect squares

$f(x)-0.875=2(x-0.25)^(2)$

$f(x)=2(x-0.25)^(2)+0.875$

the vertex is the point $(0.25,0.875)$

the axis of symmetry is

$x=0.25=(1)/(4)$

therefore

the function $f(x)=2x^(2)-x+1$ does not have a symmetry axis in $x=-(1)/(4)$

case C) $f(x)=x^(2)+2x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=x^(2)+2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+1=x^(2)+2x+1$

$f(x)+2=x^(2)+2x+1$

Rewrite as perfect squares

$f(x)+2=(x+1)^(2)$

$f(x)=(x+1)^(2)-2$

the vertex is the point $(-1,-2)$

the axis of symmetry is

$x=-1$

therefore

the function $f(x)=x^(2)+2x-1$ does not have a symmetry axis in $x=-(1)/(4)$

case D) $f(x)=x^(2)-2x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=x^(2)-2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+1=x^(2)-2x+1$

$f(x)=x^(2)-2x+1$

Rewrite as perfect squares

$f(x)=(x-1)^(2)$

the vertex is the point $(1,0)$

the axis of symmetry is

$x=1$

therefore

the function $f(x)=x^(2)-2x+1$ does not have a symmetry axis in $x=-(1)/(4)$

the answer is

$f(x)=2x^(2)+x-1$

axis of symmetry is the x value of the vertex

for
y=ax^2+bx+c
x value of vertex=-b/2a

first one
-1/2(2)=-1/4
wow, that is right

answer is first one
f(x)=2x^2+x-1

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