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Naya [18.7K]
3 years ago
12

Can someone add these up and tell me what my final grades would be

Mathematics
2 answers:
denis23 [38]3 years ago
4 0

Answer:

yea sure

Step-by-step explanation:

math pretty sure its A

civics B

LA b

pretty sure i am right

Lelu [443]3 years ago
3 0
Okaya ahahahahahaanana
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Round 2769.44 to the nearest tenths place
vlada-n [284]
It would be 2769.44 still since it would technically be 2769.440 if you were rounding to the TENTHS place. The Zero(0) doesn't change anything so the answer is

2769.44
4 0
3 years ago
Fabienne rolls a ball 72 inches across the floor. This is 9 times the distance that Olivier rolls his ball across the floor. Let
RoseWind [281]

Answer:

8 inches

Step-by-step explanation:

Oliver rolls his ball 1/9th of the distance Fabienne rolls his, so just divide 72 by 9 to get your answer.

8 0
3 years ago
Read 2 more answers
In 2015 there were 759 laptops at lightning high school starting a 2016 the school box 75 more laptops at the end of each year t
nikdorinn [45]

Answer:

Step-by-step explanation:

Given the following :

Number of laptops in 2015 = 759

t(x) = 75x + 759 ; model to determine number of laptops at the school. X years after 2015.

Number of laptops in the school in 2020:

4 0
3 years ago
Need help with Calculus 1 inverse trig functions
lidiya [134]

Answer:

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}

Where u is a function of x as provided:

y=3tan^{-1}(x+\sqrt{1+x^2})

If we set

u=(x+\sqrt{1+x^2})

Then

\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}

\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}

Taking the derivative of y

y'=3[tan^{-1}(x+\sqrt{1+x^2})]'

Using the change of variables

\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}

\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}

Operating

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}

\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}

8 0
3 years ago
Find the interquartile range (IQR) of the data in the dot plot below. chocolate chips 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10
Brums [2.3K]

*The dot plot is shown in the attachment below

Answer:

2

Step-by-step explanation:

Interquartile range is the difference between the upper median (Q3) and the lower median (Q1).

First, let's write out each value given in the data. Each dot represents a data point.

We have:

2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7

=>Find the median:

Our median is the middle value. The middle value is the 6th value = 4

==>Upper median Q3) = the middle value of the set of data we have from the median to our far right.

2, 3, 3, 4, 4, |4,| 4, 5, [5], 6, 7

Our upper median = 5

==>Lower median(Q1) = the middle value of the data set we have from our median to our far left.

2, 3, [3], 4, 4, |4,| 4, 5, 5, 6, 7

Lower median = 3

==>Interquartile range = Q3 - Q1 = 5-3 = 2

3 0
4 years ago
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