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3 years ago

This is my hardest one so far. plz help.

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

7 0

x - 27 = 62 → ( add 27 to both sides )

x - 27 + 27 = 62 + 27

x = 89

drag first box then box with x = 89

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What is the volume of a regular pyramid having a base area of 24 in.² and a height of 6 inches?

Assoli18 [71]

Answer:

48 in³

Step-by-step explanation:

The volume (V) of a pyramid is calculated as

V = $\frac{1}{3}$ × area of base × height

here area of base = 24 and height = 6, so

V = $\frac{1}{3}$ × 24 × 6 = 8 × 6 = 48 in³

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2 years ago

What solution is 2x+6=6+2x?

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Answer:

Step-by-step explanation:

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1 year ago

What is the leading coefficient of the polynomial? 7x + 5x2 − 9x3 − 10

expeople1 [14]

Answer:

The leading coefficient of the polynomial is -9.

Step-by-step explanation:

9 is the leading coefficient of this polynomial because if you rearrange the exponents from greatest to least, you will get this :

-9x3 + 5x2 + 7x -10

Your leading coefficient is the number included with your degree or highest exponent.

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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

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

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

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

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2 years ago

What is 2 and negative three using integers

Darina [25.2K]

If you mean "how to write the sum", the answer is

$2+(-3)$

The result of this sum is $-1$ , because you start from 2 on the number line and take 3 steps to the left (negative integers mean going left, positive integers mean going right)

4 0

2 years ago