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2 years ago

PLEASE HELP DIRECTIONS INCLUDED IN PHOTOS.

Mathematics

2 answers:

son4ous [18]2 years ago

5 0

I just took the test yesterday and passed
1. First one is False second one is true
2. the points are -2,-3 1,6 2,9
3. -3
4. 1

Helen [10]2 years ago

4 0

1) False, True
2) Coordinates are: (-2,0), (1,6), and (2,8)
3) -3
4) -5

-Belle

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PLS HELP

sertanlavr [38]

Answer:

C. 76 + 7i

Step-by-step explanation:

49 squared is 7 and since it's negative, it is an imaginary number. Therefore, it is 7i. Which leaves C to be the correct option.

8 0

2 years ago

Black_prince [1.1K]

Check your last post :) I answered this

3 0

2 years ago

Read 2 more answers

Which function passes through the points (2,3) and (4,4)?

Pie

A(2;3), B(4;4)
$y=ax+b$
$a=\dfrac{4-3}{4-2}=\dfrac{1}{2}$
$3=\dfrac{1}{2}*2+b$
$3=1+b$
$b=2$

$y=\dfrac{1}{2}x+2$

:)

6 0

2 years ago

You have just completed an experiment measuring the length (in mm) of 12 soybean plants after 3 days of sprouting. The measureme

sweet-ann [11.9K]

Answer:

The new IQR is 22.

Step-by-step explanation:

We are given the following data of length (in mm) of 12 soybean plants after 3 days of sprouting.

53, 47, 51, 54, 43, 39, 61, 57, 55, 46, 44, 43

Sorted data:

39, 43, 43, 44, 46, 47, 51, 53, 54, 55, 57, 61

Formula:

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median ==\dfrac{6^{th}+7^{th}}{2} \dfrac{47+51}{2} = 49$

$Q_1 =\dfrac{3^{rd}+4^{th}}{2} = \dfrac{43 + 44}{2} = 43.5\\\\Q_3 =\dfrac{9^{th}+10^{th}}{2}= \frac{54 + 55}{2} = 54.5$

IQR = $Q_3 -Q_1 =54.5-43.5= 11$

If every measurement is doubled, then, the IQR will also double itself.

Thus,

New IQR =

$2\times \text{IQR}\\=2\TIMES 11\\=22$

Thus, the new IQR is 22.

4 0

3 years ago

Please help with this Calculus questions

Triss [41]

Answer:

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\frac{193}{100}=1.93$ .

Step-by-step explanation:

To find $\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du$ .

First, calculate the corresponding indefinite integral:

Integrate term by term:

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u}} =\int{2 d u} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant rule $\int c\, du = c u$

$\int{2 d u}} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u} = {\left(2 u\right)} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$

$2 u - {\int{\frac{3 u^{9}}{2} d u}} + \int{\frac{2 u^{4}}{5} d u} = 2 u - {\left(\frac{3}{2} \int{u^{9} d u}\right)} + \left(\frac{2}{5} \int{u^{4} d u}\right)$

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$

$2 u - \frac{3}{2} {\int{u^{9} d u}} + \frac{2}{5} {\int{u^{4} d u}}=2 u - \frac{3}{2} {\frac{u^{1 + 9}}{1 + 9}}+ \frac{2}{5}{\frac{u^{1 + 4}}{1 + 4}}$

Therefore,

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u} = - \frac{3 u^{10}}{20} + \frac{2 u^{5}}{25} + 2 u = \frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)$

According to the Fundamental Theorem of Calculus, $\int_a^b F(x) dx=f(b)-f(a)$ , so just evaluate the integral at the endpoints, and that's the answer.

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}=\frac{193}{100}$

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=0$

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}-\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=\frac{193}{100}$

6 0

3 years ago