Ask question

Ask question

All categories

4 years ago

5

Work out the smallest integer value of x that satisfies the inequality 4x-3>22. ( I have not learned this question in maths )

1 answer:

matrenka [14]4 years ago

5 0

4x>22+3

4x>25

x>25/4

x>6.35

You might be interested in

Determine if each set of side lengths could be used to create a triangle. *Explain your reasoning.*

grin007 [14]

Answer:

See below

Step-by-step explanation:

By the Triangle Inequality Theorem, if given side lengths $a$ , $b$ , and $c$ , where $c$ is the longest side, then $a+b\geq c$

<u>Problem A</u>

This set of side lengths creates a triangle since $2+6\geq7$

<u>Problem B</u>

This set of side lengths creates a triangle since $3+3\geq 3$ , which is known as an equilateral triangle since all of its side lengths are equal to each other

<u>Problem C</u>

This set of side lengths creates a triangle since $5+7\geq 12$

6 0

2 years ago

What is the slope of a line that is perpendicular to 5 x – 2 y = 1 ?

lord [1]

Answer:

The slope is 5/2

4 0

3 years ago

At what point do the increasing and decreasing parts of the graphs in Part A intersect?

noname [10]

(-1,3)
Not sure if it’s right but it should be

5 0

3 years ago

HELP ASAPP!!!!!!!!!!!!!!!!

Drupady [299]

Answer:

1st one

Step-by-step explanation:

$\frac{1x}{8} + \frac{1x}{8} \\ \frac{2 x}{8} = \frac{1x}{4}$

7 0

3 years ago

Find dy/dx by implicit differentiation for ysin(y) = xcos(x)

tatyana61 [14]

Answer:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

Step-by-step explanation:

So we have:

$y\sin(y)=x\cos(x)$

And we want to find dy/dx.

So, let's take the derivative of both sides with respect to x:

$\frac{d}{dx}[y\sin(y)]=\frac{d}{dx}[x\cos(x)]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[y\sin(y)]$

We can use the product rule:

$(uv)'=u'v+uv'$

So, our derivative is:

$=\frac{d}{dx}[y]\sin(y)+y\frac{d}{dx}[\sin(y)]$

We must implicitly differentiate for y. This gives us:

$=\frac{dy}{dx}\sin(y)+y\frac{d}{dx}[\sin(y)]$

For the sin(y), we need to use the chain rule:

$u(v(x))'=u'(v(x))\cdot v'(x)$

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.

So, our derivative is:

$=\frac{dy}{dx}\sin(y)+y(\cos(y)\cdot\frac{dy}{dx}})$

Simplify:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}$

And we are done for the right.

Right Side:

We have:

$\frac{d}{dx}[x\cos(x)]$

This will be significantly easier since it's just x like normal.

Again, let's use the product rule:

$=\frac{d}{dx}[x]\cos(x)+x\frac{d}{dx}[\cos(x)]$

Differentiate:

$=\cos(x)-x\sin(x)$

So, our entire equation is:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}=\cos(x)-x\sin(x)$

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

$\frac{dy}{dx}(\sin(y)+y\cos(y))=\cos(x)-x\sin(x)$

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

And we're done!

5 0

3 years ago