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3 years ago

7. Meghan said the opposite of the sum of -12 and 4 is 8. Do you agree? Why or why not?

Mathematics

1 answer:

muminat3 years ago

8 0

I agree, the sum of -12 and 4 is -8. so, the opposite of that would be positive 8.

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How many distinguishable permutation are possible with all the letters of the word sequence

Marizza181 [45]

Step-by-step explanation:

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3 years ago

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brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

8. The loudness of a sound depends on_____. A. The width of the wave B. The brightness of the wave C. The length of a sound wave

cupoosta [38]

Answer:

Step-by-step explanation:

loudness depends on the amplitude or height of the sound waves

5 0

3 years ago

A cylinder has radius r and height h. A. How many times greater is the surface area of a cylinder when both dimensions are multi

Ierofanga [76]

Answer: A. Factor 2 => 4x greater

Factor 3 => 9x greater

Factor 5 => 25x greater

Step-by-step explanation: A. A cylinder is formed by 2 circles and a rectangle in the middle. That's why surface area is given by circumference of a circle, which is the length of the rectangle times height of the rectangle, i.e.:

A = 2.π.r.h

A cylinder of radius r and height h has area:

$A_{1}$ = 2πrh

If multiply both dimensions by a factor of 2:

$A_{2}$ = 2.π.2r.2h

$A_{2}$ = 8πrh

Comparing $A_{1}$ to $A_{2}$ :

$\frac{A_{2}}{A_{1}}$ = $\frac{8.\pi.rh}{2.\pi.rh}$ = 4

Doubling radius and height creates a surface area of a cylinder 4 times greater.

By factor 3:

$A_{3} = 2.\pi.3r.3h$

$A_{3} = 18.\pi.r.h$

Comparing areas:

$\frac{A_{3}}{A_{1}}$ = $\frac{18.\pi.r.h}{2.\pi.r.h}$ = 9

Multiplying by 3, gives an area 9 times bigger.

By factor 5:

$A_{5} = 2.\pi.5r.5h$

$A_{5} = 50.\pi.r.h$

Comparing:

$\frac{A_{5}}{A_{1}}$ = $\frac{50.\pi.r.h}{2.\pi.r.h}$ = 25

The new area is 25 times greater.

B. By analysing how many times greater and the factor that the dimensions are multiplied, you can notice the increase in area is factor². For example, when multiplied by a factor of 2, the new area is 4 times greater.

3 0

3 years ago

If anyone knows answer with steps that will be greatly appreciated :)

OLga [1]

Answer:

The area formula is= 1/2(a+b)×height

1/2×20×6=60metres squared

Step-by-step explanation:

kindly correct me if am wrong

6 0

3 years ago