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iren2701 [21]
3 years ago
15

If you can solve this question without calculator then ill give you brainliest

Mathematics
1 answer:
liq [111]3 years ago
4 0

Answer:

4580

Step-by-step explanation:

56x74=4144

382+96=478

4622-42=4580

I did the math in my head and I don't know if it is correct but I hope this helps!

Happy holidays!

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The line tangent to the graph of g(x) = x^{3}-4x+1 at the point (2, 1) is given by the formula
Usimov [2.4K]

well, about A and D, I just plugged the values on the slope formula of

\bf \begin{array}{llll} g(x)=x^3-4x+1\\ L(x) = 8(x-2)+1 \end{array} \qquad \begin{cases} x_1=1.9\\ x_2=2.1 \end{cases}\implies \cfrac{f(b)-f(a)}{b-a}

for A the values are 8.01 and 8.0, so indeed those "slopes" are close. \textit{\huge \checkmark}

for D the values are -2.25 and 8.0, so no dice on that one.

for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.

checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice. \textit{\huge \checkmark}

for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.

for E, do they intercept at (2,1)?  well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent. \textit{\huge \checkmark}

for F, we know the slope of the line L(x) is 8, is g'(2) = 8?  let's check

recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice. \textit{\huge \checkmark}

6 0
3 years ago
Mathematically verify the outlier(s) in the data set using the 1.5 rule.
statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

Divide the data set in two equal parts.

(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

Q_1=\dfrac{11+13}{2}

Q_1=\dfrac{24}{2}

Q_1=12

And

Q_3=\dfrac{16+16}{2}

Q_3=\dfrac{32}{2}

Q_3=16

The interquartile range is:

IQR=Q_3-Q_1

IQR=16-12

IQR=4

The data values lies outside the interval [Q_1-1.5IQR,Q_3+1.5IQR] are known as outliers.

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]

[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

3 0
3 years ago
Thank you for helping
omeli [17]

Answer:

12

Step-by-step explanation:

the original ratio was 3/5 , so apply that to 20 and 3/5 of 20 is 12

6 0
3 years ago
Read 2 more answers
(06.02
BARSIC [14]

Answer:

The experimental probability is larger than the relative frequency.

Step-by-step explanation:

3 0
3 years ago
You have $36 in your lunch account and plan to spend $4 each school day for lunch.
Wewaii [24]

Answer: $36/$4=9 days

Step-by-step explanation:

8 0
3 years ago
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