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3 years ago

If two halves two whole 1/2 are added to two whole 1/2 the result is

2 answers:

4 0

1/2 + 1/2= 1
1/2 + 1/2 + 1/2= 1 1/2
1/2 + 1/2 + 1/2 + 1/2= 2
1/2 + 1/2 + 1/2 + 1/2 + 1/2= 2 1/2
1/2 + 1/2 + 1/2 + 1/2+ 1/2 + 1/2= 3

Hope this helped! :)

Ivahew [28]3 years ago

3 0

1 is the answer hope this helps

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what is the equation of the quadratic function that passes through the points (0,-1),(1,0),and (2,-3)

Alexus [3.1K]

Answer:

-2x^2 + 3x -1

Step-by-step explanation:

y = ax^2 + bx + c

(0,-1):

-1 = a(0)^2 + b(0) + c

-1 = 0 + 0 + c

-1 = c

(1,0):

0 = a(1)^2 + b(1) + c

0 = a + b + c

(2,-3):

-3 = a(2)^2 + b(2) + c

-3 = 4a + 2b + c

Replace c with -1

0 = a + b -1

1 = a + b

-3 = 4a + 2b -1

-2 = 4a + 2b

-1 = 2a + b

Solve:

2a + b = -1

-1a - b = -1

a = -2

2a + b = -1

-2a -2b = -2

-b = -3 divide both by -1: b = 3

a = -2; b = 3; c = -1

y = ax^2 + bx + c or y = -2x^2 +3x -1

6 0

2 years ago

A university offers 3 calculus classes: Math 2A, 2B and 2C. In both parts, you are given data about a group of students who have

ankoles [38]

Answer:

4 students

Step-by-step explanation:

The total number of students is given by:

$|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| + |A\cap B \cap C| = N$

Where N is the total number of students, A is the number of students that have taken Math 2A, B is the number of students that have taken Math 2B, and C is the number of students that have taken Math 2C:

$|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| + |A\cap B \cap C| = N\\51+80+70-15-20-13 + |A\cap B \cap C| = 157\\ |A\cap B \cap C| =4$

4 students in Group A have taken all three classes.

*The number of students that have taken two classes has be subtracted so those students do not get counted twice

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3 years ago

. Shanice is making a garden. The area of the garden is 24 ft2. Use the following shape and information to find the base (6−4) a

crimeas [40]

Answer:

The base and height of the garden are 4.60 ft and 5.60 ft.

Step-by-step explanation:

The area of the garden is, A = 24 ft².

The base height of the garden are:

base (b) = 6x - 4

height (h) = x + 3

Compute the value of x as follows:

$A=\frac{1}{2}\times b \times h\\\\24=\frac{1}{2}\times (6x-4) \times (x+3)\\\\48=6x^{2}+18x-4x-12\\\\6x^{2}+14x-60=0\\\\3x^{2}+7x-30=0\\$

The last equation is a quadratic equation.

Compute the roots of the quadratic equation as follows:

$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }\\x = \frac{ -14 \pm \sqrt{14^2 - 4(3)(-30)}}{ 2(3) }\\x = \frac{ -14 \pm \sqrt{556}}{ 6 }\\x = \frac{ -14 }{ 6 } \pm \frac{2\sqrt{139}\, }{ 6 }\\x = -\frac{ 7}{ 3 } \pm \frac{ \sqrt{139}\, }{ 3 }\\x = 1.59661\\x\approx 1.60$