What is the answer for this

Mathematics

1 answer:

Leviafan [203]2 years ago

6 0

I think that
the triangle is 6
the square is -5
and the star is -20
but i would get a second answer bc i cant certainly say that<span />

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Suppose that y varies jointly with w and x and inversely with z and y = 32- when w = 6, x = 20, and z = 3. Write the equation th

natulia [17]

Y directly proportional to WX and inversely to Z.

Y=k(WX)/Z
Where k is the constant of proportionality.
So, the first thing is to find the value of this constant.

k = YZ/WX
= (32×3)/(6×20)
= 96/120
= 0.8
The <span>equation that models the relationship will be;
Y = 0.8(WX/Z)</span>

3 0

3 years ago

Perry borrows $350 at a simple interest rate of 4.5%. Perry pays back the loan in 30 months. How much interest does Perry pay on

Nitella [24]

Answer: the interest on the loan is $39.38

Step-by-step explanation:

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the loan.

P represents the principal or amount taken as loan

R represents interest rate

T represents the duration of the loan in years.

From the information given,

P = $350

R = 4.5%

There are 12 months in a year. Converting 30 months into years, it becomes

30/12 = 2.5. so

T = 2.5 years

Therefore

I = (350 × 4.5 × 2.5)/100

I = $39.38

3 0

3 years ago

(Geometry course question) Explain why it is not possible to construct an equilateral triangle that has

Nutka1998 [239]

Answer:

√3 is irrational

Step-by-step explanation:

The location of the third point of a triangle can be found using a rotation matrix to transform the coordinates of the given points.

<h3 /><h3>Location of point C</h3>

With reference to the attached figure, the slope of line AC is √3, an irrational number. This means the line AC <em>never passes through a point with integer coordinates</em>. (Any point with integer coordinates would be on a line with rational slope.)

<h3>Equilateral triangle</h3>

The line segments making up an equilateral triangle are separated by an angle of 60°. If two vertices are on grid squares, the third must be a rotation of one of them about the other through an angle of 60°. The rotation matrix is irrational, so the rotated point must have irrational coordinates.

The math of it is this. For rotation of (x, y) counterclockwise 60° about the origin, the transformation matrix is ...

$\left[\begin{array}{cc}\cos(60^\circ)&\sin(60^\circ)\\-\sin(60^\circ)&\cos(60^\circ)\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}x'\\y'\end{array}\right]$

Cos(60°) is rational, but sin(60°) is not. For any non-zero rational values of x and y, the sum ...

cos(60°)·x + sin(60°)·y

will be irrational.

As in the attached diagram, if one of the coordinates of the rotated point (B) is zero, then one of the coordinates of its image (C) will be rational. The other image point coordinate cannot be rational.