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3 years ago

PLEASE HELP QUICK!!!

Mathematics

1 answer:

dsp733 years ago

4 0

add me on disc its not letting me explain what i wanted to explain its Andrew#6506

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Variables and Inequalities

sleet_krkn [62]

Answer:

x>4
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hope it helps pls mark braiest

5 0

4 years ago

Using first principal find the gradient of the curve whose equation is y=x^5. show working pleas

Keith_Richards [23]

Answer:

The gradient of $y = x^{5}$ is $\frac{dy}{dx} = 5\cdot x^{4}$ .

Step-by-step explanation:

Geometrically speaking, the gradient of a curve is the slope of the tangent line at a given point of the curve. From perspective of Differential Calculus, the gradient is the first derivative of the curve. Let $y = x^{5}$ , then the gradient of the curve is:

$\frac{dy}{dx} = 5\cdot x^{4}$ (1)

The gradient of $y = x^{5}$ is $\frac{dy}{dx} = 5\cdot x^{4}$ .

6 0

3 years ago

Point R is at (2, 1.2) and Point T is at (2, 2.5) on a coordinate grid. The distance between the two points is

Sergeu [11.5K]

<h3>Answer: 1.3</h3>

========================================

Work Shown:

The two points we have are $(x_1,y_1) = (2,1.2)$ and $(x_2,y_2) = (2,2.5)$

Apply the distance formula

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(2-2)^2+(1.2-2.5)^2}\\\\d = \sqrt{(0)^2+(-1.3)^2}\\\\d = \sqrt{0+1.69}\\\\d = \sqrt{1.69}\\\\d = 1.3\\\\$

-------------------------

A shortcut is possible by subtracting the y coordinates of the two points, and making the result positive through the use of absolute value.

So we either say

|y1-y2| = |1.2-2.5| = |-1.3| = 1.3

or,

|y2 - y1| = |2.5 - 1.2| = |1.3| = 1.3

We get the same result. This shortcut is valid because the x coordinates of both points are the same.

5 0

3 years ago

Does x>x have a solution

RSB [31]

Answer:

yes

Step-by-step explanation:

3 0

4 years ago

Part D<br><br> If Tristan attaches a 4-kilogram weight to the spring, how far will it stretch?

Monica [59]

Answer:

Step-by-step explanation:

F=kx

Use this formula of Hooke's law

4 0

3 years ago

Read 2 more answers