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3 years ago

Solve the system using any method. x-3y + 2z = 4 -x+4y + 6 = 40 5x - 3y - 10z = -32

1 answer:

3 0

1. x=4+y-2z (move all terms not containing x to the right side of the equation) ; y= -4x + x + 2z (isolate) ; z= 2 - x/2 + y/2 (isolate)
2. y= 1/4x + 17/2 (slope intercept form)
3. x= -32/5 + 3y/5 + 2z ; y=32/3+5x/3-10z/3 ; z= 16/5+x/2 - 3y/10 (isolate the variable by dividing each side by factors that don’t contain variable)

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Write an expression that represents the sum of a number and 12

vodka [1.7K]

X+12=y
Hope this helps :)

5 0

3 years ago

Frank cut a pie into 10 equal slices there are three slices left what is the measure of the angle formed by the 3 slices left

viva [34]

The three remaining slices would form 108 degrees.
Hope this helped (and that I was right) =P

6 0

3 years ago

a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Ri

SSSSS [86.1K]

In the $x$ direction we consider the $m=2$ subintervals [0, 1] and [1, 2] (each with length 1), while in the $y$ direction we consider the $n=2$ subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of $R$ are (1, 0), (2, 0), (1, 2), (2, 2).

Let $f(x,y)=7x+5y^2$ . The volume of the solid is approximately

$\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}$

###

More generally, the lower-right-corner Riemann sum over $m=\mu$ and $n=\nu$ subintervals would be

$\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)$

Then taking the limits as $\mu\to\infty$ and $\nu\to\infty$ leaves us with an exact volume of $\dfrac{808}3$ .

7 0

3 years ago

Please help it’s due soon :( I’ll mark brainliest!!

Assoli18 [71]

Answer:

The answer is A. 105

Step-by-step explanation:

it's more tilted so it can't be a lower number, the only option is 105

8 0

3 years ago

Which of the following quadratic regression equations best fits the data shown below

attashe74 [19]

Let
A (-4,35)
B (-3,20)
C (-2,10)
D (-1,5)
E (0,8)
F (1,10)
G (2,15)
H (3,30)
I (4,48)

using a graph tool

case a) y=2.23x²+1.62x+5.26
see the attached figure N 1

case b) y=4.63x²+0.84x+4.14
see the attached figure N 2

case c) y=0.98x²+3.06x+6.02
see the attached figure N 3

case d) y=3.72x²+2.94x+3.21
see the attached figure N 4

using a Quadratic Regression Calculator
the Best Fit Second-Degree Quadratic Regression is
2.228x²+1.617x+5.255
see the attached figure N 5

therefore

the answer is
case a) y=2.23x²+1.62x+5.26

3 0

3 years ago

Read 2 more answers