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3 years ago

Solve x/-4 > 8 1. x > -2 2. x < -32 3. x < -2 4. x > -32

Mathematics

1 answer:

Gnom [1K]3 years ago

8 0

Answer:

x<-32

Step-by-step explanation:

I hope this helped have a good night

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The point P(5, −4) lies on the curve y = 4/(4 − x).

dalvyx [7]

Answer:

Step-by-step explanation:

Slope refers to the steepness of the line.

Slope is equal to change in value of y divided by change in value of x.

The slope of the secant line is given by $\frac{f(x)-f(a)}{x-a}$ .

Given points are $P\left ( 5,-4 \right )\,,\,Q\left ( x,\frac{4}{4-x} \right )$

So, slope of the secant line is $\frac{\frac{4}{4-x}+4}{x-5}=\frac{4+4(4-x)}{(x-5)(4-x)}=\frac{20-4x}{(x-5)(4-x)}=\frac{-4(x-5)}{(x-5)(4-x)}=\frac{-4}{4-x}=\frac{4}{x-4}$

(i) At x = 4.9,

$\frac{4}{4.9-4}=\frac{4}{0.9}=4.444444$

(ii) At x = 4.99,

$\frac{4}{4.99-4}=\frac{4}{0.99}=4.040404$

(iii) At x = 4.999,

$\frac{4}{4.999-4}=\frac{4}{0.999}=4.004004$

(iv) At x = 4.9999,

$\frac{4}{4.9999-4}=\frac{4}{0.9999}=4.0004$

(v) At x =5.1,

$\frac{4}{5.1-4}=\frac{4}{1.1}=3.636364$

(vi)

At x = 5.01,

$\frac{4}{5.01-4}=\frac{4}{1.01}=3.960396$

(vii) At x = 5.001,

$\frac{4}{5.001-4}=\frac{4}{1.001}=3.996004$

(viii) At x = 5.0001,

$\frac{4}{5.0001-4}=\frac{4}{1.0001}=3.9996$

(b)

Slope of the tangent line is 4.

5 0

3 years ago

[Help Asap, Will Mark Brainliest] Answer is in the image,

Vera_Pavlovna [14]

Z=118 as vertically opposite angles are the same

x= (8x-50)

3 0

3 years ago

Read 2 more answers

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

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3 years ago

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Answer:

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