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2 years ago

Need help! just need the answer!

Mathematics

2 answers:

Bas_tet [7]2 years ago

7 0

Answer:

Step-by-step explanation:

k=1 ,i THINK

Colt1911 [192]2 years ago

6 0

Answer is Definitely

A

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Given the function f(x) = 3x, find the value of f−1(27). f−1(27) = 1 f−1(27) = 3 f−1(27) = 9 f−1(27) = 81

Mariulka [41]

Fun fact
if f(a)=b then f⁻¹(b)=a
so
if f⁻¹(b)=a then f(a)=b

so we find f(x)=27 and the solve for x

f(x)=3x=27
3x=27
divide both sides by 3
x=9

so if f(9)=27 then f⁻¹(27)=9
problem solved

answer is 9

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3 years ago

find the slope and y-intercept of the line that is perpendicular to y= -2/3x+5 and passes through the line -6,4

nexus9112 [7]

$y = mx+b\\\\[m is the slope, b is the y-intercept]\\\\y=\frac{2}{3}x+5\\$

That means 2/3 is the slope and 5 is the y-intercept.

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

6 0

3 years ago

What is 5/10 times 2/4 show step by step explanation HELP ASAP

Darya [45]

5/10 *2/4

Cross out 2 and 10, divide by 2. 2/2=1 , 10/2=5

5/5*1/4= 5/20

reduce the fraction

divide by 5 for the numerator and denominator

5/5=1

20/5=4

5/10*2/4=1/4

Answer: 1/4

5 0

3 years ago

The two spinners below are spun. Find the probability of spinning a 4 and a C.

r-ruslan [8.4K]

Step-by-step explanation:

S = { 1, 2, 3, 4, 5, 6 7, 8 }

n ( S ) = 8

Let A be the event of getting 4,

A = { 4 }

n ( A ) = 1

P ( A )

= n ( A ) / n ( S )

= 1 / 8

Therefore, the probability of spinning a 4 is 1 / 8.

S = { A, B, A, C, A, B }

n ( S ) = 6

Let Y be the event of getting C,

Y = { C }

n ( Y ) = 1

P ( Y )

= n ( Y ) / n ( S )

= 1 / 6

Therefore, the probability of spinning a C is 1 / 6.

3 0

3 years ago

A box of donuts containing 6 maple bars, 3 chocolate donuts, and 3 custard filled donuts is sitting on a counter in a work offic

Svetlanka [38]

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

$\mathbf{Bars = 6}$

$\mathbf{Chocolate = 3}$

$\mathbf{Custard= 3}$

The total number of donuts in the box is:

$\mathbf{Total= 6 + 3 + 3}$

$\mathbf{Total= 12}$

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

$\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}$

So, we have:

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}$

Simplify

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}$

Multiply

$\mathbf{Pr = \frac{9}{132}}$

Divide

$\mathbf{Pr = 0.068}$

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068