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3 years ago

The constant value of the ratio of two proportional quantities

Mathematics

1 answer:

Kitty [74]3 years ago

5 0

Think it of a fraction problem,

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Which expression shows 0.95 in expanded form?

raketka [301]

In standard form,it should be 0 +0.9 +0.05

4 0

3 years ago

PLEASE HELP NOW!!!!!!!!!!!!!!!!!!!

Dahasolnce [82]

Answer:

8 + k² - 6 > 8(k + 2) - 4k when k = 7

Step-by-step explanation:

8 + k² - 6 = 8 + 7² - 6 = 51

8(k + 2) - 4k = 8(7 + 2) - 4(7) = 8(9) - 28 = 44

8 + k² - 6 > 8(k + 2) - 4k when k = 7

8 0

3 years ago

An expression that is equivalent to (-8a-1)-(-7a+5)

horrorfan [7]

Answer:

Step-by-step explanation:

(-8a - 1)- (-7a + 5) = -8a - 1 - 7a*(-1) + 5*(-1) { (-1) is disturbed to all the terms in (-7a +5)}

= -8a - 1 + 7a - 5

= -8a + 7a -1 - 5 {Combine like terms}

= -a - 6

5 0

3 years ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

4 years ago

I need help on this! ASAP

strojnjashka [21]

Answer:

-2

Step-by-step explanation:

Good luck! :)

3 0

2 years ago

Read 2 more answers