Identify the constant of proportionality in the equation. 3y = 15x A) 1 B) 3 C) 5 D) 15

Mathematics

2 answers:

umka21 [38]3 years ago

6 0

Answer: The answer is c)5

sveticcg [70]3 years ago

3 0

Answer:

c) 5

Step-by-step explanation:

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3 years ago

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Mazyrski [523]

Your answer will be the third option.

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3 years ago

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25 POINTS AND BRAINLIEST PLEASE HELP ASAP

dexar [7]

M is a midpoint of BC so:

$M=\left(\dfrac{\boxed{2}\boxed{a}+a}{\boxed{2}},\dfrac{\boxed{0}+b}{2}\right)=\left(\dfrac{\boxed{3}\boxed{a}}{\boxed{2}},\dfrac{\boxed{b}}{\boxed{2}}\right)$

Length of MA:

$MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=
\sqrt{\left(\dfrac{\boxed{3}a}{\boxed{2}}\right)^2+\left(\dfrac{b}{2}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

Length of NB:

$NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\
\sqrt{\left(\dfrac{-3a}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

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3 years ago

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Viefleur [7K]

The outer exponent is 4 so there will be 5 terms all together. By rules of exponents

C(4,0)(3^4)(2^0)(x^8)(y^0) + C(4,1)(3^3)(2^1)(x^6)(y^3) + C(4,2)(3^2)(2^2)(x^4)(y^6) + C(4,3)(3^1)(2^3)(x^2)(y^9) + C(4,4)(3^0)(2^4)(x^0)(y^12)

The coefficient of the 3rd term: C(4,2)(3^2)(2^2) = 6 x 9 x 4 = 216

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3 years ago

Sec^6x(secxtanx)-sec^4(secxtanx)=sec^5xtan^3x<br> Verify the trigonometric identity

mario62 [17]

I guess you mean

$\sec^6x(\sec x\tan x)-\sec^4x(\sec x\tan x)=\sec^5x\tan^3x$