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lianna [129]
3 years ago
13

2/3 divided by 3/4 simplified

Mathematics
2 answers:
Hatshy [7]3 years ago
7 0

Answer: 8/9 = 0.8888889

mariarad [96]3 years ago
6 0
8/9 is the answer to your question
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there are 25 students in math class. yesterday 6students were absent .what percent of the students were absent
kodGreya [7K]
Divide the number absent by the total and multiply by 100
P=6/25 x 100% = 24%
4 0
4 years ago
Is 2 + 2 = 4 - 1 is 3 quick maths I need to know ASAP
Kitty [74]
Yes, 2 + 2 = 4 - 1 does equal 3 
6 0
3 years ago
Read 2 more answers
Determine the number of real solutions each quadratic equation has. y = 12x2 - 9x 4 real solution(s) 10x y = -x2 2 real solution
nordsb [41]

All the given equations have 2 real solutions except for equation 4 which has only one real root.

<h3>What is a quadratic equation?</h3>

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is  ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

The given quadratic equation

y = 12x^2 - 9x+ 4

where, a = 12, b = -9 and c = 4

x = -b + \sqrt{b^2 - 4ac} /2a

x = -(-9)+ \sqrt{(-9)^2 - 4(12)(4)} /2(12)

x = 1.063 or -0.313

Second Equation

10x +y = -x^2 +2 \\y = x^{2} +10x - 2

where, a = 1, b = 10, c = -2

substitute the values into the equation

x = -b + \sqrt{b^2 - 4ac} /2a

-10 + \sqrt{10^2 - 4(1)(-2)} /2(1)

x = 0.195 or -10.195

Third Equation

4y - 7 = 5x^2 - x +2 +3y

y = 5x^{2} -x -5

where, a = 5, b = -1, c = - 5

x = -b + \sqrt{b^2 - 4ac} /2a

-(-1) + \sqrt{(-1)^2 - 4(5)(-5)} /2(5)

x = 0.905 or -0.55

Fourth Equation

y = (-x +4)^2 \\y = x^{2} -8x +16

where, a = 1, b = -8, c = 16

x = -b + \sqrt{b^2 - 4ac} /2a

x = -(-8) + \sqrt{(-8)^2 - 4(1)(16)} /2(1)

x = 4

Learn more about quadratic equations;

brainly.com/question/13197897

6 0
2 years ago
Show that Circle R with center (2, -9) and radius 5 is similar to Circle Q with center (-4, 3) and radius 3 by using a series of
faltersainse [42]

Answer:

It can be proved that the circle R is similar to the circle Q by translating the circle R a displacement of (-6, 12).

Step-by-step explanation:

We can demonstrate that Circle R is similar to Circle Q by translating the center of the former one to the center of latter one. Meaning that every point of circle R experiments the same translation. Vectorially speaking, a translation is defined by:

O'(x,y) = O(x,y) + T(x,y) (1)

Where:

O(x,y) - Original point.

O'(x,y) - Translated point.

T(x,y) - Translation vector.

If we know that O(x,y) = (2,-9) and O'(x,y) = (-4,3), then the translation vector is:

T(x,y) = O'(x,y)-O(x,y)

T(x,y) = (-4,3) - (2,-9)

T(x,y) = (-6,12)

It can be proved that the circle R is similar to the circle Q by translating the circle R a displacement of (-6, 12).

4 0
3 years ago
you and a friend decided to take a road trip to somewhere in Pennsylvania on the weekend. You are traveling at an average speed
Zinaida [17]
62 multiplied by 3 is 186
Add on 31 for the half hour
 and you get 217
8 0
3 years ago
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