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2 years ago

HELP ME YOULL GET 25 POINTS!!!!!!!!

Mathematics

1 answer:

tester [92]2 years ago

8 0

Answer:

Step-by-step explanation:

This graph is digits going by eigths.

For 2 1/2, mark a dot right between the 2 and 3, on the fourth tick mark.

For -3/8, go left of 0 3 tick marks. (Right between 0 and -1).

For -1 3/4, go right of -2 by 2 tick marks. (Right between -2 and -1)

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The table shows the results of drawing letter tiles from the bag. What is the probability that the next tile drawn will have the

Orlov [11]

The answer is B i think

8 0

3 years ago

PLEASE HELP RIGHT AWAY

Angelina_Jolie [31]

Answer:

12 flips

Step-by-step explanation:

4 0

2 years ago

Taletha is preparing for an exam in Algebra II. She knows she will be expected to determine if two functions are inverses of eac

seropon [69]

Answer:

f ○ g(x) = x and g ○ f(x) = x is the correct composition

Step-by-step explanation:

INVERSE FUNCTIONS: Two functions are said to be inverse of each other if for every y = f(x), there exists a function g(x),

such that x = $f^{-1} (y)$ = g(y)

Now, here if we need to show tha two functions f(x) and g(x) are inverse functions of each other , then show that for every x:

1. f o g(x) = f(g(x))= x

2. g o f(x) = g(f(x)) = x

If any two functions satisfy these two conditions, then they are inverse of each other.

5 0

3 years ago

The heat developed in an electric wire varies jointly as the wires resistance, the time the current flows, & the square of t

never [62]

Answer:

4 ohms

Step-by-step explanation:

The question is "In two minutes a current of 5 amps develops 1,200 heat units in a wire of 8 ohms resistance. What resistance does a similar wire have, which develops 6,000 heat units with a current of 10 amps in 5 minutes?"

Current, I₁ = 5 A

Heat, Q₁ = 1200 units

Resistance, R₁ = 8 ohms

Time, t₁ = 2 min = 120 s

New heat, Q₂ = 6000 units

Current, I₂ = 10 A

New time, t₂ = 5 min = 300 s

We need to find the new resistance.

Heat developed is given by :

$Q=I^2Rt$

$\dfrac{Q_1}{I_1^2R_1t_1}=\dfrac{Q_2}{I_2^2R_2t_2}\\\\R_2=\dfrac{Q_2\times I_1^2R_1t_1}{Q_1I_2^2t_2}\\\\R_2=\dfrac{6000\times 5^2\times 8\times 120}{1200\times 10^2\times 300}\\\\=4\ \Omega$

So, the new resistance is 4 ohms.

5 0

2 years ago

Solve the equation 2cosA = 3tanA

zavuch27 [327]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\
-----------------------------\\\\

2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}
\\\\\\
2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)
\\\\\\
2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\
0=[2sin(A)-1][sin(A)+2]\implies 
\begin{cases}
0=2sin(A)-1\\
1=2sin(A)\\
\frac{1}{2}=sin(A)\\\\
sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\
\frac{\pi }{6},\frac{5\pi }{6}\\
----------\\
0=sin(A)+2\\
-2=sin(A)
\end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A

8 0

2 years ago