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Drupady [299]
4 years ago
5

I already drew it out can you please tell me if you think ur is unique if so why?

Mathematics
1 answer:
WINSTONCH [101]4 years ago
6 0
I literally had the almost the exact same question I couldn't find it either imma tell you when I get it ;)
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Siri round each number to the place of the underlined digit 7.158 The underlined number is five
monitta
The answer is 7.160 !!
6 0
4 years ago
How do you figure out this equation?
maw [93]
Equation of a circle is (x-x1)+(y-y1)=r²
when x1 and y1 are the coordinates of the centre

So you can substitute the centre in to get (x-5)²+(y+9)²=r²

When you draw a diagram it's obvious the radius is 9

9²=81

Therefore equation is (x-5)²+(y+9)²=81

Any questions, just ask :)

6 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!
svet-max [94.6K]

Answer:

tan (C) = 2.05

Step-by-step explanation:

Given:

A right angled triangle CDE right angled at ∠D.

Side CD = 39

Side DE = 80

Side CE = 89

We know, from trigonometric ratios that, the tangent of any angle is equal to the ratio of the opposite side to the angle and the adjacent side of the angle.

Therefore, tangent of angle C is given as:

\tan(\angle C)=\frac{DE}{CD}

Plug in the given values and solve for angle C.This gives,

\tan(\angle C)=\frac{80}{39}\\\\\tan(\angle C)=2.051\approx 2.05(Rounded\ to\ nearest\ hundredth)

Therefore, the measure of tangent of angle C is 2.05.

4 0
3 years ago
Let y represent the total cost of publishing a book (in dollars). Let x represent the number of the copies of the book printed.
Oduvanchick [21]
Im sorry idk this answer

6 0
3 years ago
The miles-per-gallon rating of passenger cars is a normally distributed random variable with a mean of 33.8 mpg and a standard d
EleoNora [17]

Answer:

The probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

Step-by-step explanation:

Let the random variable <em>X</em> represent the miles-per-gallon rating of passenger cars.

It is provided that X\sim N(\mu=33.8,\ \sigma^{2}=3.5^{2}).

Compute the probability that a randomly selected passenger car gets more than 37.3 mpg as follows:

P(X>37.3)=P(\frac{X-\mu}{\sigma}>\frac{37.3-33.8}{3.5})

                   =P(Z>1)\\\\=1-P(Z

Thus, the probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

7 0
3 years ago
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