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Marianna [84]
1 year ago
8

Is y=2x-1 proportional

Mathematics
1 answer:
GREYUIT [131]1 year ago
7 0
Since y=2x+1 does not pass through the origin (i.e. there exists the constant term of 1 ), the variables y and x are not directly proportional - though, they do have a linear relationship.
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Pls hurry! Will make brainliest!! What is the formula that relates circumference and diameter?
Westkost [7]

Answer:

Circumference = πd

where d = diameter

Step-by-step explanation:

When using this formula you don't need to multiply by 2 like you do when using the radius because the diameter is twice the radius of a circle.

3 0
2 years ago
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Find the distance between the two points.<br> (-6, -7) and (154, - 43)
Vitek1552 [10]

Answer:

(160, 36)

Step-by-step explanation:

Distance is always positive because, well, you aren't -50 meters away from your house or something. You are just 50 meters away from your house. And 160 is the distance between -6 and 154, because there is one minus and one plus, you add them, because that is how far they are apart, and then-43 and -7 were easy just subtract them.

8 0
2 years ago
Simplity
Vikentia [17]

Answer:

A.

5 ^ (5/6)

Step-by-step explanation:

5 ^ (1/2)  * 5 ^ (1/3)

We know that a^b * a^ c = a ^ (b+c)

5 ^ (1/2 + 1/3)

5^ (3/6+2/6)

5^ (5/6)

8 0
3 years ago
Read 2 more answers
Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

3 0
2 years ago
In the number 2145857, how does the digit 5 in the thousands place compare to the digit 5 in the tens place?
fgiga [73]
It is 1,000 bigger than the one in the tens place. 5,000 compared to 50 
7 0
3 years ago
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