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4 years ago

Helpppp Plsssss Asap!! Show your work

Mathematics

2 answers:

satela [25.4K]4 years ago

8 0

Answer:

Step-by-step explanation:

The anwser is C

the inverse of 5x is x/5 and the inverse of +4 is -4 so the answer would be x-4/5

solmaris [256]4 years ago

7 0

Answer:

B. f-1(x) = (x - 4)/5.

Step-by-step explanation:

Let y = 5x + 4

5x = y - 4

x = (y - 4)/5

f-1(x) = (x - 4)/5.

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Please help!! I only need the even numbers!! Show work!!

stepan [7]

2=-2(-2)+b

2=4+b

-2=b

y=-2x-2

4=-3(-5)+b

4=15+b

-11=b

y=-3x-11

-5=-3/2(1)+b

-5=-3/2+b

b=-7/2 or -3 1/2

y=-3/2-7/2

8 0

3 years ago

How many solutions are possible for a triangle with A = 113° , a = 15, and b = 8

statuscvo [17]

Answer:

One solution.

Step-by-step explanation:

To determine the number of possible solutions for a triangle with A = 113° , a = 15, and b = 8, we're going to use the law of sines which states that: "When we divide side a by the sine of angle A it is equal to side b divided by the sine of angle B, and also equal to side c divided by the sine of angle C".

Using the law of sines we have:

$\frac{sin(A)}{a} = \frac{sin(B)}{b}$

$\frac{sin(113)}{15} = \frac{sin(B)}{8}$

Solving for B, we have:

$sin(B)=0.4909$

∠B = 29.4°

Therefore, the measure of the third angle is: ∠C = 37.6°

There is another angle whose sine is 0.4909 which is 180° - 29.4° = 150.6 degrees. Given that the sum of all three angles of any triangle must be equal to 180 deg, we can't have a triangle with angle B=113° and C=150.6°, because B+C>180.

Therefore, there is one triangle that satisfies the conditions.

7 0

3 years ago

The integers $r$ and $k$ are randomly selected, where $-3 < r < 6$ and $1 < k < 8$. what is the probability that the

e-lub [12.9K]

Denote by $R,K$ the random variables representing the integer values $r,k$ , respectively. Then $R\sim\mathrm{Unif}(-2,5)$ and $K\sim\mathrm{Unif}(2,7)$ , where $\mathrm{Unif}(a,b)$ denotes the discrete uniform distribution over the interval $[a,b]$ . So $R$ and $K$ have probability mass functions

$p_R(r)=\begin{cases}\dfrac18&\text{for }r\in\{-2,-1,\ldots,5\}\\\\0&\text{otherwise}\end{cases}$

$p_K(k)=\begin{cases}\dfrac16&\text{for }k\in\{2,3,\ldots7\}\\\\0&\text{otherwise}\end{cases}$

We want to find $P\left(\dfrac RK\equiv0\pmod n\right)$ , where $n$ is any integer.

We have six possible choices for $K$ :

(i) if $K=2$ , then $\dfrac RK$ is an integer when $R=\pm2,0,4$ ;

(ii) if $K=3$ , then $\dfrac RK$ is an integer when $R=0,3$ ;

(iii) if $K=4$ , then $\dfrac RK$ is an integer when $R=0,4$ ;

(iv) if $K=5$ , then $\dfrac RK$ is an integer when $R=0,5$ ;

(v) if $K=6$ or $K=7$ , then $\dfrac RK$ is an integer only when $R=0$ in both cases.

If the selection of $R,K$ are made independently, then the joint distribution is the product of the marginal distribution, i.e.

$p_{R,K}(r,k)=p_R(r)\cdot p_K(k)=\begin{cases}\dfrac1{48}&\text{for }(r,k)\in[-2,5]\times[2,7]\\\\0&\text{otherwise}\end{cases}$

That is, there are 48 possible events in the sample space. We counted 12 possible outcomes in which $\dfrac RK$ is an integer, so the probability of this happening is $\dfrac{12}{48}=\dfrac14$ .