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4 years ago

Drag each number to show whether or not it is a solution to the inequality shown. Inequality: 20≥5+3

Mathematics

2 answers:

Bond [772]4 years ago

8 0

It’s not a solution because 5+3 = 8 and 20 is greater than eight but there’s a line under the greater sign which means it’s greater than or equal to and 20 is not equal to 8 (if that makes sense)

posledela4 years ago

3 0

Hope this helps! Was gonna add an in-depth answer... but I think the image is enough. Just substitute x for any of these numbers when doing the equations and you’ll see how the answers are less than, equal to, or more than 20. :]

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Felix was given $125 for his birthday from his grandmother. He plans to put this money toward the purchase of a $250 skateboard.

defon

Its is B! like I just worked it out.

4 0

3 years ago

I know the converse but I need to know which ones are true

Ray Of Light [21]

The converse of the statement would be

"If x ≤ 7, then x² ≤ 49"

but this is false. Take x = -8; while it's true that -8 ≤ 7, the second inequality is not, since (-8)² = 64 is not smaller than 49.

4 0

3 years ago

How to solve part ii and iii

iragen [17]

(i) Given that

$\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}$

when $x=1$ this reduces to

$\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$2 \tan^{-1}(y) = \dfrac\pi3$

$\tan^{-1}(y) = \dfrac\pi6$

$\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)$

$\implies \boxed{y = \dfrac1{\sqrt3}}$

(ii) Differentiate $\tan^{-1}(xy)$ implicitly with respect to $x$ . By the chain and product rules,

$\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}$

(iii) Differentiating both sides of the given equation leads to

$\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0$

where we use the result from (ii) for the derivative of $\tan^{-1}(xy)$ .

Solve for $\frac{dy}{dx}$ :

$\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0$

$\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)$

$\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}$

From part (i), we have $x=1$ and $y=\frac1{\sqrt3}$ , and substituting these leads to

$\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}$

$\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}$

$\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}$

as required.

3 0

2 years ago

What is 5.37 as a mixed number.

Gnom [1K]

Answer: 5.37 as a mixed number is 5 37/100

To write a decimal number as a mixed number, you must separate the whole part of the decimal part:
5.37 = 5 + 0.37 = 5 0.37
And express the decimal part as a fraction:
0.37=37/100
Then:
Answer: 5.37 as a mixed number is 5 37/100

7 0

3 years ago

A dolphin jumps from the water at a initial velocity of 16 feet per second the equation h=-8t^2 + 16t models the dolphins height

german

Answer:

The answer to the equation -8t^2+16t = -48

Step-by-step explanation:

Add ''-8t^2'' to ''16t'' and the answer you get is ''-48''

6 0

3 years ago

Read 2 more answers