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3 years ago

Find the absolute Value of -0.99

Mathematics

2 answers:

Leona [35]3 years ago

7 0

Answer:

0.99

Step-by-step explanation:

That's the distance from 0.

KengaRu [80]3 years ago

3 0

Answer:

.99

Step-by-step explanation:

it is point .99 units away from the 0 on the number line.

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You have a dance class every three days. Today is Monday, and you have dance class. In how many more days will you have dance cl

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7 days in a week, 3 days a rotation.

multiples of 7: 7, 14, 21, 28

multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24

the LCM is 21, because it is the smallest multiple of both numbers. there will be 21 more days until you take dance class on Mondays

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McNeary brought home a weekly pay check with these deductions: Federal--$45.67, FICA-- 24.63, state--19.04. How much does he pay

vlada-n [284]

McNeary gets payed annually about;
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A scuba driver is investigating a coral reef 6 meters below sea. If he descends another 21 meters what will be his final locatio

tatiyna

Answer:

27 meters

Step-by-step explanation:

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The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

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3 years ago

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Answer: I meant to comment.

Step-by-step explanation: sorry

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