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3 years ago

Find the location of the absolute maximum and absolute minimum of the function on the interval [ -4,0).

1 answer:

N76 [4]3 years ago

3 0

Maximum is 0 and minimum is -4. Due to the brackets, that means that this is where the value terminates. The parentheses next to the zero means equal to

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For each equation choose a value for x and then solve

ss7ja [257]

Answer:

Step-by-step explanation:

For each equation choose a value for x and then solve to find the corresponding y value that makes that equation true.

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2 years ago

What is equivalent to 8/9 divided by 3/4

Pavel [41]

Answer:

Dividing by a fraction is the same as multiplying with the opposite of the fraction. 8/9 / 3/4=8/9×4/3=32/27=1 5/27.

Step-by-step explanation:

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2 years ago

What is the inverse of the function f(x) = 2x + 1?

BaLLatris [955]

Answer:

$f^{-1} = \frac{x-1}{2}$

Step-by-step explanation:

$f(x) = 2x+1$

Replace it with y

$y = 2x+1$

Exchange the values of x and y

$x = 2y+1$

Solve for y

$x = 2y+1$

Subtracting 1 from both sides

$2y = x-1$

Dividing both sides by 2

$y = \frac{x-1}{2}$

Replace it by $f^{-1}$

So,

$f^{-1} = \frac{x-1}{2}$

3 0

3 years ago

Read 2 more answers

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

3 0

3 years ago

Read 2 more answers

Please answer this correctly

Lubov Fominskaja [6]

The awnser to this is 2 and also 3

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3 years ago