How many different ways can you add two odd numbers to get a sum of 20 and add examples

A student claims that the rectangular point below can be represented by the

Sladkaya [172]

Since we don't have a figure we'll assume one of them is right and we're just being asked to check if they're the same number. I like writing polar coordinates with a P in front to remind me.

It's surely false if that's really a 3π/7; I'll guess that's a typo that's really 3π/4.

P(6√2, 7π/4) = ( 6√2 cos 7π/4, 6√2 sin 7π/4 )

P(-6√2, 3π/4) = ( -6√2 cos 3π/4, -6√2 sin 3π/4 )

That's true since when we add pi to an angle it negates both the sine and the cosine,

cos(7π/4) = cos(π + 3π/4) = -cos(3π/4)

sin(7π/4) = sin(π + 3π/4) = -sin(3π/4)

Answer: TRUE

3 0

3 years ago

Solve the system by the elimination method.

NikAS [45]

2x+3y-10=0 (1)
+
4x-3y-2=0 (2)
____________
6x -12=0 (if you just want the resulting equation. It is 6x-12=0)
x=2
take x=2 and put it into equation (2)
4(2) -3y -2=0
-3y= 2-8
y= 2
(x=2,y=2)

5 0

3 years ago

Given the polynomial 21x 4 + 3y - 6x 2 + 34 What polynomial must be subtracted from it to obtain 29x 2 - 7 ? What polynomial mus

fredd [130]

Answer:

1. 21x⁴+3y-35x² + 41

2. -21x⁴-3y+6x² + x

Step-by-step explanation:

When adding and subtracting polynomials , you can use the distributive property to add or subtract the coefficients of like terms.

1. The polynomial is 21x⁴ + 3y -6x² + 34

To obtain polynomial 29x² -7 , we must subtract some polynomial from it.

Let that polynomial be k.

So, 21x⁴ + 3y -6x² + 34 - k = 29x² -7

k = 21x⁴ + 3y - 6x² +34 - 29x² +7 = 21x⁴ + 3y - 35x² + 41

2. To obtain a first degree polynomial, let that polynomial be x +34

So, 21x⁴ + 3y - 6x² + 34 + K = x + 34

K = x + 34 - 21x⁴ -3y + 6x² - 34

= -21x⁴ - 3y + 6x² + x

6 0

3 years ago

What Is 1+1=? Help!!!

mestny [16]

2!!! I better get 5 points

4 0

3 years ago

Read 2 more answers

Find the lengths and slopes of the diagonals to name the parallelogram. Choose the most specific name. E (-2, -4), F(0, -1), G(-

Kay [80]

Answer:

1) d) Square

2) Proofs that PWRS is a rhombus are

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Step-by-step explanation:

The given points (x, y) of the parallelogram are;

E(-2, -4), F(0, -1), G(-3, 1), H(-5, -2)

The slope, m, of the segments are found as follows;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

By computation, the slope of segment EF = 1.5

The slope of segment FG = -0.67

The slope of segment GH = 1.5

The slope of segment HE = -0.67

Therefore, EF is parallel to GH and FG is parallel to HE

The length of the sides are;

$\sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

By computation, the length of segment EF = 3.61

The length of segment FG = 3.61

The length of segment GH = 3.61

The length of segment HE = 3.61

The diagonals are;

EG and FH

The length of segment EG = 5.099

The length of segment FH = 5.099

Therefore, the diagonals are equal and the parallelogram is a square

2) The given dimensions are;

P(-1, 3), Q(-2, 5), R(0, 4), S(1, 2)

A rhombus has all sides equal

The length of segment PQ = 2.24

The length of segment QR = 2.24

The length of segment RS = 2.24

The length of segment PS = 2.24

The diagonals are;

QS and PR

The length of segment QS = 4.24

The length of segment PR = 1.41

The slope of segment QR = -0.5

The slope of segment PS = -0.5

The slope of segment RS = -2

The slope of segment QP = -2

Therefore, QS≠QR the parallelogram is a rhombus

The correct option ;

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Where there are acute angles in parallelogram PQRS, then the correct option is d) Length of QR and PS is 2.2 and Length of RS and QP is 2.2

8 0

3 years ago