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3 years ago

What is the x-intercept for y=7/4x+1/4

Mathematics

1 answer:

Brums [2.3K]3 years ago

4 0

9514 1404 393

Answer:

-1/7

Step-by-step explanation:

To find the x-intercept, set y=0 and solve for x.

0 = 7/4x +1/4

Multiply by 4/7

0 = x + 1/7

Subtract 1/7

-1/7 = x

The x-intercept is -1/7.

_____

In decimal, that is -0.142857... (6-digit repeating decimal)

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Evaluate the following expression using the values given.

mel-nik [20]

Answer:

-19

Step-by-step explanation:

2×2-2z4+y2-x2+z4

now we put the value,

4-2(2to the power 4)+(3 to the power 2)-(-4 to the power 2)+(2 to the power 4)

4-2×16+9-16+16

4-32+9

4+9-32

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3 years ago

What is an equation of the line that passes through the point (−5,−1) and is parallel to the line 4x+5y=45?

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This gives you, 4x + 5y = -25 as your final answer.

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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:

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

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