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3 years ago

The absolute ______ of 2 is 2. 1.) absolute value 2.)sign 3.)opposite

Mathematics

2 answers:

Paul [167]3 years ago

8 0

The absolute value of 2 is 2

tino4ka555 [31]3 years ago

4 0

1.) Absolute value

Hope this helps

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Can someone help me with this problem please!! i’ll give brainliest!

Sphinxa [80]

Answer:

its b i think dont trust though

Step-by-step explanation:

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3 years ago

How do I find the indicated derivative?

sergey [27]

Multiply the fraction by both things in the parenthesis.

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3 years ago

Find the quadratic function that fits the following data. which one function fits.

Oksana_A [137]

Answer:

Step-by-step explanation:

The general rule for the quadratic function is

$y=ax^2+bx+c$

Use the data from the table:

$y(50)=130\Rightarrow 130=a\cdot 50^2+b\cdot 50+c\\ \\y(70)=130\Rightarrow 130=a\cdot 70^2+b\cdot 70+c\\ \\y(90)=200\Rightarrow 200=a\cdot 90^2+b\cdot 90+c$

We get the system of three equations:

$\left\{\begin{array}{l}2500a+50b+c=130\\ \\4900a+70b+c=130\\ \\8100a+90b+c=200\end{array}\right.$

Subtract these equations:

$\left\{\begin{array}{l}4900a+70b+c-2500a-50b-c=130-130\\ \\8100a+90b+c-2500a-50b-c=200-130\end{array}\right.\Rightarrow \left\{\begin{array}{l}2400a+20b=0\\ \\5600a+40b=70\end{array}\right.$

From the first equation

$b=-120a$

Substitute it into the second equation:

$5600a+40\cdot (-120a)=70\Rightarrow 800a=70,\\ \\ a=\dfrac{7}{80},\\ \\ b=-120\cdot \dfrac{7}{80}=-\dfrac{21}{2}=-10.5$

So,

$2500\cdot \dfrac{7}{80}+50\cdot (-10.5)+c=130\Rightarrow 218.75-525+c=130\\ \\c=130-218.75+525=436.25$

The quadratic function is

$y=\dfrac{7}{80}x^2-10.5x+436.25\\ \\y=0.0875x^2-10.5x+436.25$

4 0

3 years ago

What is the area of the kite 10 12 12 18

Brums [2.3K]

P=2A
q so it’s 18
That is the answe

8 0

3 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago