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Lady bird [3.3K]
3 years ago
11

Jackie won tickets playing the bowling game at the local arcade. The first time, she won 60 tickets. The second time, she won a

bonus, which was 4 times the number of tickets of the second prize. Altogether she won 200 tickets. How many tickets was the second prize?
Mathematics
1 answer:
Fynjy0 [20]3 years ago
4 0

Answer: i'm pretty sure she got 35 the second time. hope this helps

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Murljashka [212]

Answer:

5/7

Step-by-step explanation:

we find the greatest number that can be multiplied to get each number. it is 11. 11 goes into 55 5 times, and into 77 7 times therefore giving us 5/7

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3 years ago
PLEASE HELP!!!! WILL MARK BRAINLIEST IF CORRECT!!!!!!
PilotLPTM [1.2K]
-2 would be your answer
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3 years ago
At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not
Nikitich [7]
If f(x, y, z) = c represent a family of surfaces for different values of the constant c. The gradient of the function f defined as \nabla f is a vector normal to the surface f(x, y, z) = c.

Given <span>the paraboloid

y = x^2 + z^2.

We can rewrite it as a scalar value function f as follows:

f(x,y,z)=x^2-y+z^2=0

The normal to the </span><span>paraboloid at any point is given by:

\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\  \\ =2xi-j+2zk

Also, the normal to the given plane 3x + 2y + 7z = 2 is given by:

3i+2j+7k

Equating the two normal vectors, we have:
</span>
2x=3\Rightarrow x= \frac{3}{2}  \\  \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}

Since, -1 = 2 is not possible, therefore there exist no such point <span>on the paraboloid y = x^2 + z^2 such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2</span>.
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3 years ago
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Andreyy89
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3 years ago
‼️ help with all these questions 20pts + brainliest ‼️
eimsori [14]

Answer:

1) Slope-intercept form

2) 9200

3) 2 months

4) (0,200)

Step-by-step explanation:

A shelter had 200 animals in foster homes at the beginning of spring and the number of animals in foster homes at the end of the summer could be represented by  

y=3000x+200 ............ (1)

Where x is the number of months and y is the number of animals.

1) The equation (1) is written in the slope-intercept form of a straight line equation.

2) After 3 months means x = 3 and the number of animals in the foster home after 3 months will be (3000 ×3 + 200) = 9200 (Answer)

3) Let after x months the animal population will become 6200.

So, 6200 = 3000x + 200

⇒ 3000x = 6000

⇒ x = 2 months (Answer)  

4) If we put x = 0 in equation (1), then we get y = 200.

So, (0,200) is a point on the graph of the line. (Answer)

5 0
3 years ago
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