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Vinil7 [7]
3 years ago
6

What is the answer please help me no Links

Mathematics
1 answer:
mihalych1998 [28]3 years ago
8 0
I’m pretty sure the answer is y=50m+25
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What are the answers to:<br> 1. -4s = -16<br> 2. 63 = -9d<br> 3. h/-3 = 12<br> 4. g/12 = -10
UkoKoshka [18]
The answer for the first one would be 4 just divide both 16 & 4.

The answer for the second one would be -7 just divide both 63 & -9.

The answer for the third one would be -36 just multiply both 12 and -3.

The answer for the fourth one would be -120 just multiply -10 and 12.
7 0
3 years ago
What is the fraction 70/75 equivalent to?
ki77a [65]

Answer:14/15

Step-by-step explanation:

7 0
3 years ago
Can someone help me please ill give brainy
Gnom [1K]
Y = 6

Since there is no slope the equation would look like this y=0x+6

Well 0 times x is 0 so y=+6
6 0
3 years ago
Read 2 more answers
Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,
Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j}  = {(1,0), (0,1)}

is a base in \mathbb{R}^2

and F is linear, then

<em>{F(1,0), F(0,1)}  </em>

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0
4 years ago
X = ? y = ? 16 45 degrees
mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}

For x, we will be using the Cosine Function:

\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}(16)Cosine(45^{\circ})\text{ =  x}(16)(\frac{1}{\sqrt[]{2}})\text{ = x}\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}\text{ 8}\sqrt[]{2}\text{ = x}

Therefore, x = 8√2.

For y, we will be using the Sine Function.

\text{  Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}\text{ (16)Sine }(45^{\circ})\text{ =  y}\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}\text{ 8}\sqrt[]{2}\text{ = y}

Therefore, y = 8√2.

5 0
2 years ago
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