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Arlecino [84]
3 years ago
9

What are the solutions to the system of equations? y=−x^2−5x−6 x+y=−3

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer:

The solutions are the points (-3,0) and (-1,-2)

Step-by-step explanation:

we have

y=-x^{2} -5x-6 -----> equation A

x+y=-3 -----> equation B

Solve the system of equations by graphing

The solution of the system of equations are the intersection points both graphs

The solutions are the points (-3,0) and (-1,-2)

see the attached figure

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-3 = 7 - BLANK pls tell me what blank is
liberstina [14]

Answer:

10

Step-by-step explanation:

-3 = 7 - x

Add x to both sides

x -3 = 7 - x +x

x - 3 = 7

Now, add 3 to both sides

x - 3 + 3 = 7 + 3

x = 10

4 0
3 years ago
Read 2 more answers
Write an equation of a line in slope-intercept from with the given slope, slope:–3/4, y-intercept: –2
larisa [96]

Answer: y=-3/4x-2

Step-by-step explanation:

Write an equation of a line in slope-intercept from with the given slope, slope:–3/4, y-intercept: –2

The y-intercept is at  (0,-2)

Then we write the equation using y=mx+b becuase that is the point slope form.

m=-3/4 which was given in the question

because y-intercept is at (0,-2)

y=-2  

x=0

-2=-3/4(0)+b

-2=0+b

-2=b

now we know that m is -3/4 and b is -2

Hence the equation is y=-3/4x-2

7 0
3 years ago
The mass of the Earth is 5.972x10^24 kg. the mass of the moon is 7.3 x 10^22kg. find the total mass, in kg, of the Earth and the
torisob [31]

Answer:

6045000000000000000000000 kg.

Step-by-step explanation:

We have been given that the mass of Earth is 5.972\times10^{24} kg. The mass of the Moon is 7.3\times10^{22} kg.

To find the total mass we will add mass of Earth and Moon.

First of all let us convert the given masses in standard form.

5.972\times10^{24}=5972000000000000000000000  

7.3\times10^{22}=73000000000000000000000

5.972\times10^{24}+7.3\times10^{22}  

5972000000000000000000000+73000000000000000000000

6045000000000000000000000

Therefore, the mass of Earth and Moon is 6045000000000000000000000 kg.

8 0
3 years ago
A lamina with constant density rho(x, y) = rho occupies the given region. Find the moments of inertia Ix and Iy and the radii of
jenyasd209 [6]

Answer:

Ix = Iy = \frac{ρπR^{4} }{16}

Radius of gyration x = y =  \frac{R}{4}

Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

Mass of disk = ρπR2

Moment of inertia about its perpendicular axis is \frac{MR^{2} }{2}. Moment of inertia of quarter disk about its perpendicular is \frac{MR^{2} }{8}.

Now using perpendicular axis theorem, Ix = Iy = \frac{MR^{2} }{16} = \frac{ρπR^{4} }{16}.

For Radius of gyration K, equate MK2 = MR2/16, K= R/4.

3 0
3 years ago
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Ludmilka [50]
.4 is smallest, 1/2 , then 60% is greatest.
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