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1 year ago

RATING BRAINLIEST Which of the following demonstrates the Distributive Property?

Mathematics

2 answers:

svetoff [14.1K]1 year ago

8 0

Answer:

just saying this so other can get brainliest

Step-by-step explanation:

ra1l [238]1 year ago

4 0

Answer:

D: 4(2a+3)=8a+3

Step-by-step explanation:

4x2a = 8a and 4x3 = 12

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What is 3.75x-5-8.75+4.5+0.5 I need help ASAP it is a simplify the expression question

miv72 [106K]

Answer:

3.75x-8.75 or

Step-by-step explanation:

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2 years ago

How do you find the area of a triangle

motikmotik

A=1/2b(h) this is the formula to find the area of a triangle

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2 years ago

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BRAINLIEST BRAINLIEST!!! EASY MATH! Do with explanation!!

Anuta_ua [19.1K]

Answer:

Answer is given below with explanations.,

Step-by-step explanation:

$1)formula \: for \: linear \: equation \: is \: \\ ax + by + c = 0 \\ 2) \: solution \: of \: the \: equation \: y = 9x - 3 \\ (2,15) \: is \: the \: solution \: \\ on \: substitutuing \: the \: points \\ 15 = 9(2) - 3 \\ 15 = 18 - 3 \\ 15 = 15 \\ hence \: proved\\ 3) \: given \:( k,7) \: is \: a \: solution\\ \: of \: y = - 4x + 1 \\ on \: substitutuing \: the \: points \\ 7 = - 4k + 1 \\ 4k = - 7 + 1 \\ 4k = - 6 \\ k = \frac{ - 6}{4} \\ k = \frac{ - 2}{3}$

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2 years ago

Eighty eighths in cups

Tamiku [17]

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3 years ago

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If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

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3 years ago