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3 years ago

Find the missing side lengths. Can someone please help???

Mathematics

1 answer:

adoni [48]3 years ago

6 0

Answer:

x = 16

y = 13.8

Step-by-step explanation:

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7x+9y=-41 and x+3y=1. Solve system of equations

galben [10]

9514 1404 393

Answer:

(x, y) = (-11, 4)

Step-by-step explanation:

You can subtract 3 times the second equation from the first to eliminate y.

(7x +9y) -3(x +3y) = (-41) -3(1)

4x = -44

x = -11

Substituting into the second equation gives ...

-11 +3y = 1

3y = 12

y = 4

The solution is (x, y) = (-11, 4).

7 0

3 years ago

General Mills, the manufacturer of Lucky Charms, claims that their new boxes hold 25% more cereal. If their original box held 50

shutvik [7]

Answer:

625

Step-by-step explanation:

basically, half of 500 is 250 halve that you get a quater of 500 (25%)

500+125=625

6 0

3 years ago

A weight of 270 pounds falls

beks73 [17]

Answer:

240

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Pleaseee someone help!! giving 20 points

choli [55]

Answer:

m=125*n

Step-by-step explanation:

for every minute that passes 125 meals are packed so you multiply 125 by the amount of minutes(n) to get m

5 0

3 years ago

Read 2 more answers

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

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

6 0

4 years ago