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alina1380 [7]
3 years ago
10

Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!!

Mathematics
1 answer:
Sholpan [36]3 years ago
7 0
X•X= x^2, 4•2=8, 8x^2
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Please help ASAP!! will give brainliest :)​
tino4ka555 [31]

Answer:

it is the first one

y=\sqrt{2x-6-1}

if i can get brainliest that would be great

4 0
3 years ago
Find the x- and y-intercepts for the linear equation.<br> – 3x + 5y = 15
Paladinen [302]

Answer:

x-intercept: (-5,0)

y-intercept: (0,3)

Step-by-step explanation:

y-intercept: when x = 0

-3(0) + 5y = 15

5y = 15

y = 3

x-intercept: when y = 0

-3x + 5(0) = 15

-3x = 15

x = -5

6 0
3 years ago
Read 2 more answers
What’s (3b-5c)(2x+5y) squared
lukranit [14]
Answer: In your problem, you said that they are squared. However, it is not written as squared. We just have to multiply them.

We can use the FOIL method.

(3b - 5c)(3x + 5y)

6bx + 15by - 10cx - 25cy

There are no like terms, so that expression is our final answer.
8 0
3 years ago
This is due in one hour
UNO [17]

Answer:

The answer to the first one is y = 4/7 x − 13 /7

The second one is already in standard form

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Three listening stations located at (3300, 0), (3300, 1100), and (-3300, 0) monitor an explosion. The last two stations detect t
erastovalidia [21]

Answer:

The coordinates of the explosion is (3300, -2750)

Step-by-step explanation:

I have plot the points and attached to thus answer for easy understanding.

Now, from the question, since station A is the first to hear the explosion, we'll make it the foci of the parabola in the graph I attached and it will be horizontal since the distance between station C and A is much more than that between station B and A. Thus, the reason why station C will have to be the other foci with the hyperbola centred at the origin.

Now, sound travels at a speed of 1100 ft/s and station B is located 1100 ft from station A. Thus, the explosion would likely have occurred at a point on the line x = 3300ft . Since station A is 3300ft from centre C = 3300,hence C² = 3300² = 10,890,000. Since it takes 4 seconds longer for the sound to reach station C than A, the sound has traveled 4(1100)= 4400 ft.

Thus, 4400 = d1 = d2 = 2a

So,2a = 4400 and so, a =2200

a² = 2200² = 4,840,000 where d1 is the distance from station C to the explosion and d2 is the distance from station A to the explosion. To find b², let's use the equation ;

c² = a² + b² and so; b² = c² - a² = 10,890,000 - 4,840,000 = 6,050,000

Equation of hyperbola is given as;

(x²/a²) - (y²/b²) = 1

Plugging in the values of a² and b², we obtain ;

(x²/4,840,000) - (y²/6,050,000) = 1

Since we have deduced that the explosion must occur on the line x= 3300, we'll put in 3300 for x to obtain ;

(3300²/4,840,000) - (y²/6,050,000) = 1

2.25 - 1 = (y²/6,050,000)

y² = (1.25 x 6,050,000)

y² = 7562500

y = √7562500

y = ± 2750

Due to the fact that the explosion will occur at a point further from station B than from station A, the explosion will take place in quadrant 4. Thus, we will take the negative value of y which is - 2750.

So explosion will occur at the coordinate (3300, -2750)

7 0
3 years ago
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