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3 years ago

BRAINLIEST + POINTS! EASY PLEASE EXPLAIN!

Mathematics

2 answers:

dmitriy555 [2]3 years ago

6 0

An upper extreme is the highest number on the number line give. So the Upper extreme is 11

kvv77 [185]3 years ago

5 0

The upper extreme is 10

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Find A/// help please!! ❤️❤️

andre [41]

It’s number 12!!!!! I know it already got answered

5 0

3 years ago

Please answer !!!!!!!

Akimi4 [234]

The measure of all the angles is given below:

angle 1= 30

angle 2= 150

angle 3= 30

angle 4= 150

angle 5= 30

angle 6= 150

angle 7= 30

angle 8= 73.5

<h3>What is angle?</h3>

An angle is formed when two straight lines or rays meet at a common endpoint. The common point of contact is called the vertex of an angle.

As,

angle 1= 30 (Vertically opposite angle)

angle 7= 30 (corresponding angle)

angle 4+30= 180 (linear pair)

angle 4= 150

angle 4= angle 2= 150 (Vertically opposite angle)

angle 2= angle 6= 150 (corresponding angle)

angle 1= angle 5= 30 (corresponding angle)

angle (2x+3)= angle 4 (corresponding angle)

2x+3= 150

x= 73.5

Lean more about angle here:

brainly.com/question/13954458

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7 0

2 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Solve by elimination <br> X - 6y + 2z = 5<br> 2x-3y+z=4<br> 3x + 4y - z = -2

Vesnalui [34]

Answer:

x=1

y=-3

z=-7

Step-by-step explanation:

6 0

3 years ago

Find the median, lower quartile, and upper quartile for the set of data.

harkovskaia [24]

$\text{Find the median:}\\\\\text{To find the median, you would cross out the numbers from the side}\\\text{until you're left with the remainders. Put them in order and solve}\\\\1, 2, 4 ,5, 6, 7, 8, 9, 10, 11\\\text{Add 6 and 7 then divide by 2}\\\\6+7=13\\\\11\div2=6.5\\\\\boxed{\text{The median is 6.5}\\}\\\\\text{The lower quartile is the median of the numbers less than the median, so}\\\text{the numbers from 1-6}\\\\\text{When you cross out the numbers, you would be left with 4}$

$\boxed{\text{The lower quartile is 4}}\\\\\text{Do the same for the upper quartile for the numbers greater than half}\\\text{of the data set}\\\\\boxed{\text{Your upper quartile would be 9}}$

8 0

3 years ago

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