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Sergeeva-Olga [200]
3 years ago
9

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

Mathematics
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

<em>Replace it with y</em>

y = 2x+1

<em>Exchange the values of  x and y</em>

x = 2y+1

<em>Solve for y</em>

x = 2y+1

<em>Subtracting 1 from both sides</em>

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<em>Dividing both sides by 2</em>

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<em>Replace it by </em>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
7 0
3 years ago
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