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3 years ago

Solve by combining equations PLS HELP! 3x - 2y = 12 -3x+8y= -6

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

3 0

3x−2y=12;−3x+8y=−6

Step: Solve3x−2y=12for x:

3x−2y=12

3x−2y+2y=12+2y(Add 2y to both sides)

3x=2y+12

2y+12

(Divide both sides by 3)

y+4

Step: Substitute

y+4forxin−3x+8y=−6:

−3x+8y=−6

−3(

y+4)+8y=−6

6y−12=−6(Simplify both sides of the equation)

6y−12+12=−6+12(Add 12 to both sides)

6y=6

(Divide both sides by 6)

y=1

Step: Substitute1foryinx=

y+4:

y+4

(1)+4

(Simplify both sides of the equation)

Answer:

and y=1

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When solving a system of two linear equations

soldier1979 [14.2K]

Answer:

When both equations have the same slope, but not the same y-intercept, they'll be parallel to each other and no intersections means no solutions. When both equations have different slopes than regardless of the y-intercept they'll intersect for certain, therefore it has exactly one solution.

Step-by-step explanation:

Got this from google hope it helps

7 0

3 years ago

From the set {1, 3, 5}, use substitution to determine which value of x makes the equation true.

mixer [17]

{1,3,5}
ie, sub 1, 3 and 5 for x and see which one is true

20-2(1)=10
20-2=10
18=10
false

20-2(3)=10
20-6=10
14=10
false

20-2(5)=10
20-10=10
10=10

true

answer is 5

5 0

3 years ago

What is the surface area of the rectangular pyramid below?

Sidana [21]

Answer:

Option (A). 363 units²

Step-by-step explanation:

Surface area of the rectangular pyramid = Area of the rectangular base + Area of the lateral sides of the pyramid

Area of the rectangular base = Length × Width

= 11 × 11

= 121 units²

Area of the lateral sides = 4(Area of one lateral side)

= 4[ $\frac{1}{2}(\text{Base})(\text{Height})$ ]

= 2(11)(11)

= 242 square units

Area of the rectangular pyramid = 121 + 242

= 363 square units

Therefore, Option (A) will be the answer.

3 0

3 years ago

Evaluate 3x^2 - 2x + 23 if x = 0.

ale4655 [162]

[ Answer ]

[ Explanation ]

3x ^ 2

3 * 0 = 0

0 ^ 2 = 0

0 - 2x

2 * 0 = 0

0 - 0 = 0

0 + 23 = 23

<> Eclipsed <>

4 0

4 years ago

PLEASE HELP! See attachment:

beks73 [17]

$\bf f(x)=x^2e^{-2x}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=2xe^{-2x}+x^2\cdot -2e^{-2x}\implies \cfrac{dy}{dx}=xe^{-2x}(2-2x)
\\\\\\
\cfrac{dy}{dx}=\cfrac{x(2-2x)}{e^{2x}}\implies 0=\cfrac{x(2-2x)}{e^{2x}}
\\\\\\
0=x(2-2x)\implies 
\begin{cases}
0=x\\
0=2-2x\implies 2x=2\implies x=1
\end{cases}$

now, bear in mind, that zeroing out the denominator, also gives critical points, usually asymptotic points, where the derivative is undefined, now, in this case, the denominator is never zero, so we don't get any from the denominator, just from the numerator, and are 0 and 1

now check the picture below

running a first-derivative test on it, those are the values on those regions

you get a negative, regardless of what it might be, what matters is the sign
you get a positive, and then a negative

so, f(x) goes down, then up then down

now, you can see, there's on relative minimum and a relative maximum

6 0

3 years ago