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2 years ago

The Constant of Proportionality from the following equation is: y= 7/9x

Mathematics

2 answers:

Ilia_Sergeevich [38]2 years ago

5 0

Answer:

k = $\frac{7}{9}$

Step-by-step explanation:

The equation of proportionality is

y = kx ← k is the constant of proportionality

y = $\frac{7}{9}$ x ← is in this form

with k = $\frac{7}{9}$

Andrej [43]2 years ago

3 0

Answer:

$\frac{7}{9}$

Step-by-step explanation:

A direct variation function is in this general form:

$y = kx\\\rule{150}{0.5}\\k -\text{ Constant of Proportionality}$

Since $\frac{7}{9}$ is in 'k's place, it is the Constant of Proportionality.

Hope this helps you.

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gayaneshka [121]

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3 years ago

Read 2 more answers

6x +8=5x=4 what is the value of x

Free_Kalibri [48]

Answer:

x = -4/11

Step-by-step explanation:

add the x values.

11x+8=4

subtract 8 from each sides

11x=-4

divide by 11

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8 0

2 years ago

(4x + 8x - 5) + (3x2 + 6x - 7)

ch4aika [34]

Answer:

7x^2 + 14x - 12

Step-by-step explanation:

Add like terms:

(4x + 8x - 5) + (3x2 + 6x - 7) becomes:

4x^2 + 8x - 5 or (grouping like terms in columns):

+ 3x^2 + 6x - 7

---------------------

7x^2 + 14x - 12

5 0

3 years ago

If tge angle at the center is 62° and the radius is 12 cm.. find the length of arc

insens350 [35]

Answer:

Step-by-step explanation:

r = 12 cm

Theta = 62

Length of arc = $\frac{theta}{360}*(2\pi r)$

$=\frac{62}{360}*2*3.14*12$

= 12.98 cm

6 0

2 years ago

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0

3 years ago