Drag each interval to a box to show us the function shown is increasing decreasing or neither over that interval

Mathematics

1 answer:

kramer3 years ago

4 0

Answer:

Increasing : -10<x<-8, -2<x<0

Decreasing : 3<x<7

Neither Increasing or Decreasing : -8<x<-5, 8<x<10

Step-by-step explanation:

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Find the difference: 7 7/8 - 3 1/4=?*

sveticcg [70]

4 5/8
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6 0

3 years ago

5+7k>=-30 and -4k +4>20

Slav-nsk [51]

Answer:

$\large\boxed{-5\leq k$

Step-by-step explanation:

$5+7k\geq-30\ \wedge\ -4k+4>20\\\\5+7k\geq-30\qquad\text{subtract 5 from both sides}\\7k\geq-35\qquad\text{divide both sides by 7}\\\boxed{k\geq-5}\\\\-4k+4>20\qquad\text{subtract 4 from both sides}\\-4k>16\qquad\text{change the signs}\\4k$

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1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

<u>Step-by-step explanation:</u>

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$