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NemiM [27]
3 years ago
13

The data in the table below for my function. What values in the table would change the relation to NOT be a function?

Mathematics
1 answer:
melamori03 [73]3 years ago
8 0
The answer is C I think
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Common factors of 28 and 40
Nady [450]
28: 1, 2, 4, 7, 14, 28

40: 1, 2, 4, 5, 8, 10, 20, 40

The common factors are 1, 2, and 4.

You can use a factor rainbow to help you find the factors (:
7 0
3 years ago
Plz help i need to find slope here
Aleksandr-060686 [28]
The answer is d which is 6/5
3 0
3 years ago
How many pounds are equal to 400 ounces
Stella [2.4K]

Answer: 25 pounds

Step-by-step explanation:

1 pound = 16 oz

400 oz / 16 oz = 25 pounds

I am not a professional, I am simply using prior knowledge!

Note- It would mean the world to me if you could mark me brainliest!

6 0
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>.
4 0
3 years ago
In ΔIJK, k = 57 inches, i = 37 inches and ∠J=141°. Find ∠I, to the nearest degree.
Sonbull [250]

Answer:

<I= 15degrees

Step-by-step explanation:

Using the cosine rule formulae;

j² = i²+k²-2i cos <J

j² = 37²+57² - 2(37)(57)cos <141

j² = 1369+ 3249- 4218cos <141

j² = 4618- 4218cos <141

j² = 4618-(-3,278)

j²= 7,896

j = √7,896

j = 88.86inches

Next is to get <I

i² = j²+k²-2jk cos <I

37² = 88.86²+57² - 2(88.86)(57)cos <I

1369 = 7,896.0996+ 3249- 10,130.04cos <I

1369 = 11,145.0996 - 10,130.04cos <I

1369 - 11,145.0996 = - 10,130.04cos <I

-9,776.0996=- 10,130.04cos <I

cos <I =9,776.0996 /10,130.04

cos<I = 0.96506

<I = 15.19

<I= 15degrees

8 0
3 years ago
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