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3 years ago

The sum of a number and twenty is greater than four times the number decreased by one solve the inequality

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

4 0

I think it's
$x \leqslant 5$
I could be wrong but I think I'm right

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1. Which of the following is the sale price for an item with an original price of $8.95 after it has been discounted 40%?

dsp73

Hey there!

1, 8.95*.4= 3.58 was take off. 8.95-3.58= 5.37
$5.37

2.<span>A change from 85 to 72 represents a negative change (decrease) of -15.294117647058824%</span>

8 0

2 years ago

8^10 x 8^4 simplified write your answer as a power

boyakko [2]

Answer: 8^14

8^10 x 8^4= 8^14

6 0

3 years ago

Read 2 more answers

The perimeters of the square and the triangle shown below are equal. What is the value of x?

Rudiy27

Perimeter is the sum of all the sides. So we can set up an equation:

$\sf 3x+1+3x+1+3x+1+3x+1=2x+8+x+8+4x+2$

Now solve for 'x', combine like terms:

When it comes to terms with variables it's just like normal addition but we keep the variable:

$\sf 3x+3x+3x+3x=12x$
$\sf 2x+x+4x=7x$

So we have:

$\sf 12x+1+1+1+1=7x+8+8+2$

Add:

$\sf 12x+4=7x+18$

Subtract 7x to both sides:

$\sf 5x+4=18$

Subtract 4 to both sides:

$\sf 5x=14$

Divide 5 to both sides:

$\boxed{\sf x=2.8}$

8 0

3 years ago

A cube with side length 4 centimeters is 25% full of sand. What is the volume of sand

earnstyle [38]

First, find the volume of the whole cube
Volume of cube = 4 x 4 x 4 = 64 cm³

The volume of sand is 25% of the volume of cube so,
Volume of sand = 25/100 x 64 = 16 cm³

4 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago