MATHEMATICS HIGH SCHOOL

Find the midpoint between the points (5, 8) and (2, 7) on the coordinate plane.a) (7/2, 15/2)
b) (7/2, 15)
c) (7, 15)
d) (6, 5)

Answers

Answer 1
Answer: midpoint = (5+2/2 , 8+7/2)
 = (7/2, 15/2)

a. (7/2, 15/2)

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The quotient of a number and 3, increased by 4 is 5, what is the expression
Solve for Q: (5q-3n)/m=f

A spinner is divide into 10 equal sections numbered from 1 to 10. You SPIN the spinner once. What is P(5)? A.)1/3
B.)1/2
C.)1/10
D.)1/4

Answers

The probability of spinning a 5 would be C. 1/10. Since none of the numbers repeat, you have an equal chance at spinning any of the numbers... 1/10.

Answer: B: P(not 1)= 10%

Step-by-step explanation:

i put C and it was wrong. showed me B is correct

Y-1=5/6(x-4) in standard form

Answers

           y - 1 = ⁵/₆(x - 4)
           y - 1 = ⁵/₆(x) - ⁵/₆(4)
           y - 1 = ⁵/₆x - 3¹/₃
             + 1         + 1
                 y = ⁵/₆x - 2¹/₃
        ⁻⁵/₆x + y = ⁵/₆x - ⁵/₆x - 2¹/₃
   -6(⁻⁵/₆x + y) = -6(-2¹/₃)
-6(⁻⁵/₆x) - 6(y) = 14
         5x - 6y = 14
If you would like to write the equation Y - 1 = 5/6 * (x - 4) in standard form, you can do this using the following steps:

Y - 1 = 5/6 * (x - 4)
Y - 1 = 5/6 * x - 5/6 * 4
Y - 1 = 5/6 * x - 10/3
Y - 5/6 * x = 1 - 10/3
Y - 5/6 * x = - 7/3       /*6
6 * Y - 5 * x = - 14

The correct result would be 6 * Y - 5 * x = - 14.

The square root of 49 equals

Answers

The square root of 49 is 7, because 7 x 7 = 49.

A square root is a number multiplied by itself to equal a certain number.
The square root of 49 is 7:)
Your welcome:)

Can anyone help? i dont understand math :(

Answers

Answer:

Step-by-step explanation:

g(4)= (√25 - 4^2)/ 4 = (√25 - 16)/4 = √9/4 = 3/4

g(-3) = (√25 -(-3)^2)/-3 = (√(25-9)/-3 = (√(16)/-3 = -4/3

h(3) = 30(1.2)^3 = 51.84

h(0)= 30(1.2)^0 = 30(1) =30

Answer:

geogrpahy

Step-by-step explanation:

h = 37

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

so

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

Solve for n. Round to the tenths place, if necessary. 9.75/24 = 5.85/n

A. n = 10.6 ounces
B. n = 14.4 ounces
C. n = 9.8 ounces
D. n = 15.2 ounces

Answers

n = 14.4 

so it would be "B" 

24 x 5.85 = 140.4 

9.75 x n = 9.75n

140.4/9.75 = 14.4 
n=14.4
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