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2 years ago

Which system is equivalent y=-2x^2 y=x-2

Mathematics

2 answers:

german2 years ago

7 0

The wording of your question suggests that there were answer choices. Mind sharing them?

The system y=-2x^2 y=x-2 would be best written as:

y=-2x^2
y=x-2 If you subst. x-2 for y in the first equation, you'll get:
x-2 = -2x^2, or 2x^2 + x - 2 = 0.

Travka [436]2 years ago

3 0

Answer:

Step-by-step explanation:

$y= -2x^{2}\\and\\y =x-2$

substituting y =x-2 in equation first we get

$y= -2x^{2}$

$(x-2)= -2x^{2}$

$2x^{2}+x-2=0$

solving it using quadratic formula we get

x =-1.281 and corresponding y =-3.281

for x =0.781 we have y =-1.219

more over we can graph both given and find the points of intersection from it.

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Willie and his brothers went to TGI Mondays. Their order consisted of 3 cheeseburgers at $8.75 each, a fish sandwich for $6.75,

Sladkaya [172]

Answer:$68.69

Step-by-step explanation:

Calculation of bill value

Cheeseburgers = 26.25

(8.75 x 3)

Fish Sandwich = 6.75

Pizza = 14.99

Sweet Tea

(4 x 2.09) = 8.36

Total of items = 56.35

Sales tax on food = 3.381

(56.35 x 6%)

Total bill value = 59.731

Calculation of tips

Bill value x 15%

=59.731 x 15%

Tips paid=8.95965

Calculation of meals cost

Total meal cost = Bill value + tips

=59.731+8.95965

=$68.69

Total meal cost is $68.69

7 0

2 years ago

Two friends decide to go on a four hour ride on off road vehicles through mountainous terrain the graph represents their elevati

Aleksandr-060686 [28]

Answer:

Extrema: relative minimum (1.25,-3.25), relative maximums (3.25,10) and (0,0)

Zeros: (0,0), (2,0), and (4,0)

End behavior: As x approaches infinity, the function approaches negative infinity. As x approaches negative infinity, the function approaches negative infinity.

Intervals of increase and decrease: increasing on (-∞, 0) and (1.25, 3.25), decreasing on (0, 1.25) and (3.25, ∞)

Positive and negative intervals: positive on (2, 4), negative on (-∞, 0), (0, 2), and (4, ∞)

plato answer

Step-by-step explanation:

8 0

2 years ago

How can the triangles be proven similar by the SSS similarity theorem? Show that the ratios StartFraction U V Over X Y EndFracti

dangina [55]

Answer:

option 1

Step-by-step explanation:

The question is as shown in the attached figure.

From the figure, we can deduce that:

∠U ≅ ∠X and ∠W ≅ ∠Z and ∠V ≅ ∠Y

So, ΔUWV similar to ΔXZY

But it is required to know " How can the triangles be proven similar by the SSS similarity theorem? "

So, we need to prove the corresponding angles are proportion

With the help of the previous corresponding angles.

So, we need to show that the ratios $\frac{UV}{XY} \ , \ \frac{WU}{ZX} \ and \ \frac{WV}{ZY}$ are equivalent.

So, the answer is option 1

5 0

2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

Segment AB falls on line 2x − 4y = 8. Segment CD falls on line 4x + 2y = 8. What is true about segments AB and CD? They are perp

Alisiya [41]

Answer:

They are perpendicular because they have slopes that are opposite reciprocals of −2 and one half.

Step-by-step explanation:

This is because x = -2 and half of -2 is 1

when we use CD line and x2 we find 8x+4y=16 when added to 2x -4y=8 would equal 10x+4y = 2 1/2 xy = 16

When we use for AB line we see they are perpendicular 2 1/2 x 2 = 5 -4y = 8 shows y to be -2 and the 1/2 line leaves -2 1/2 and x also is 2 1/2.

6 0

1 year ago

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