All categories

3 years ago

What other point will appear on the line representing this proportional relationship where the constant of proportionality is 2?

Mathematics

1 answer:

Bumek [7]3 years ago

4 0

Answer:

Step-by-step explanation:

The equation of proportionality is

y = kx ← k is the constant of proportionality

Here k = 2 , then

y = 2x

This means that the y- coordinate is twice the x- coordinate

The only point where this is true is the point

(2, 4 ) → B

You might be interested in

Find each missing length, make sure you give both exact AND approximate. Brainliest + 50pts

goldenfox [79]

B²=15²-7²
B²=225-49
B²=176
B=13.26(exact)
B=13.3(approx)

B²=17²-7²
B²=289-49
B²=240
B=15.49(exact)
B=15.5

H²=20²+14²
H²=400+196
H²=596
H=24.41(approx)
H=24.4(exact)

B²=6²-3²
B²=36-9
B²=27
B=5.196(exact)
B=5.2(approx)

3 0

3 years ago

HELPPPPP!!!!!!!!!!!!!!!!!!

diamong [38]

Answer:

I think b im not too sure

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Carolyn is making a quilt using blocks like the one shown below. All the shaded squares are the same size and all the shaded tri

dangina [55]

The answer is 117 in squared

4 0

4 years ago

Read 2 more answers

The following set of coordinates most specifically represents which figure? (−5, 6), (−1, 8), (3, 6), (−1, 4) (6 points) Paralle

Vladimir [108]

Answer:

Rhombus

Step-by-step explanation:

The given points are A(−5, 6), B(−1, 8), C(3, 6), D(−1, 4).

We use the distance formula to find the length of AB.

$|AB|=\sqrt{(-1--5)^2+(8-6)^2}$

$|AB|=\sqrt{16+4}$

$|AB|=\sqrt{20}$

The length of AD is

$|AD|=\sqrt{(-1--5)^2+(6-4)^2}$

$|AD|=\sqrt{16+4}$

$|AD|=\sqrt{20}$

The length of BC is:

$|BC|=\sqrt{(-1-3)^2+(8-6)^2}$

$|BC|=\sqrt{16+4}$

$|BC|=\sqrt{20}$

The length of CD is

$|CD|=\sqrt{(-1-3)^2+(6-4)^2}$

$|CD|=\sqrt{16+4}$

$|CD|=\sqrt{20}$

Since all sides are congruent the quadrilateral could be a rhombus or a square.

Slope of AB $=\frac{8-6}{-1--5}=\frac{1}{2}$

Slope of BC $=\frac{8-6}{-1-3}=-\frac{1}{2}$

Since the slopes of the adjacent sides are not negative reciprocals of each other, the quadrilateral cannot be a square. It is a rhombus

7 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%2081" id="TexFormula1" title=" {x}^{2} + 81" alt=" {x}^{2} + 8

borishaifa [10]

X to the power of two+81

5 0

3 years ago