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kupik [55]
2 years ago
15

Find the inverse of the function. f(x) = -3x + 7

Mathematics
1 answer:
LUCKY_DIMON [66]2 years ago
8 0
F(x) = -3x + 7
y = -3x + 7
x = -3y + 7
-3y + 7 = x
-3y = x - 7
y = -1/3x + 7/3
f^-1(x) = -1/3x + 7/3
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I need help plz help
amid [387]

We know

\boxed{\sf cos\Theta=\dfrac{b}{h}}

\\ \sf\longmapsto cos22=\dfrac{x}{47}

\\ \sf\longmapsto 0.9=\dfrac{x}{47}

\\ \sf\longmapsto x=47(0.9)

\\ \sf\longmapsto x=42.3

7 0
3 years ago
16.The least common denominator of 1/2, 1/8, and 1/10 isO A. 10.OB. 80.C. 40.D. 16.nMark for review Awill be biabliabted on the
tangare [24]

Given

The numbers, 1/2, 1/8, and 1/10.

To find: The least common denominator.

Explanation:

It is given that,

1/2, 1/8, and 1/10 .

Then,

Therefore, the LCM is,

\begin{gathered} LCM=2\times2\times2\times5 \\ =40 \end{gathered}

Hence, the least common denominator is 40.

6 0
1 year ago
A square banquet hall has a 3600m2 area. What is the length of the room?
Assoli18 [71]
The answer should be 10.
8 0
3 years ago
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Which pair of numbers is relatively prime
mote1985 [20]

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3 years ago
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Use a truth table to show that P Qand (~PV Q) A (~QV P) are equivalen
kati45 [8]

Answer:  The given logical equivalence is proved below.

Step-by-step explanation:  We are given to use truth tables to show the following logical equivalence :

P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P)

We know that

two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.

The truth table is as follows :

P     Q      ∼P     ∼Q     P⇔ Q    ∼P ∨ Q     ∼Q ∨ P        (∼P ∨ Q)∧(∼Q ∨ P)

T     T         F        F             T            T                   T                       T

T     F         F        T             F             F                   T                       F

F     T         T        F             F            T                   F                       F

F     F         T        T             T            T                   T                       T

Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.

Thus, P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P).

8 0
3 years ago
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