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Vikki [24]
3 years ago
15

HELP WITH THIS PROBLEM RIGHT NOW. If you find x first and then y, will the answer be the same or different

Mathematics
2 answers:
Alja [10]3 years ago
5 0

Answer: it will be the same!

kari74 [83]3 years ago
4 0
It would be the same
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Select all the situations for which only zero or positive solutions make sense.
san4es73 [151]

Answer:

Step-by-step explanation:

On this one you would eliminate the ones that are negative or that go into a negative. Which would be A,C, and E. The rest that are left are zero or positive. B for example is a zero because eventually there will be no candle. D would be the same in some way, the students will eventually leave the school. Finally F, the temperature would be increasing.

Answer - B,D,F

8 0
3 years ago
How many fluorine atoms are found in 1.92 grams of AsF3?
11111nata11111 [884]
First you need to find out how many moles of AsF3 are in 1.92 grams.  
The molar mass of AsF3 is 131.918 g/mole so  
1.92 g * (1 mole)/(131.918g) = .0146 moles AsF3 


 Each mole of AsF3 has 6.02 x10^23 molecules of AsF3 and each molecule of AsF3 has 3 flourines, so you need to convert from moles to molecules AsF3 to atoms of F 
 0.146 moles AsF3 * (6.02 x 10^23 molecules AsF3)/(1 mole AsF3) * (3 atoms F)/(1 molecule AsF3) = 2.64 x 10^22 atoms F
6 0
4 years ago
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
Please help me with this question
saul85 [17]
Given: 
ML = 2.3
Angle L = 70 degrees

What we want to find:
MN = x

Use the sine rule
Make sure your calculator is in degree mode

sin(angle) = opposite/hypotenuse
sin(70) = x/2.3
2.3*sin(70) = x
x = 2.3*sin(70)
x = 2.16129302780759
x = 2.2

-------------------------------
Answer: 2.2

7 0
3 years ago
Represent each of these relations on {1, 2, 3, 4} with a matrix (with the elements of this set listed in increasing order). a) {
Irina-Kira [14]

Answer:

for a  a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]

Step-by-step explanation:

in matrix, arrays are placed in rows , which represents the horizontal sides from left to right, while arrays in the column are placed vertically from top to bottom. Here, we placed the arrays in a 4x4 matrix

for a  a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]

7 0
4 years ago
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