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3 years ago

12

Help me please, 1. And 2.

1 answer:

harkovskaia [24]3 years ago

4 0

Answer:

Step-by-step explanation:YOU HAVE TO SHOW ME THE WHOLE THING SO THAT I CAN ANSWER IT DUH

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Mary needed capital for her bakery. She borrowed $60 000 for 4 years at a simple interest rate of 8% per year. How much money wi

vlada-n [284]

Answer:

$ 79600

Step-by-step explanation:

interest=prt (p is the amount borrowed, r is the rate, t is the time)

p=60000 , r=8%=0.08 , t=4 years

interest=60000*0.08*4

A=19600

the amount paid at the end of 4 years: 60000+19600=$79600

4 0

4 years ago

Nigel flips a coin 50 times and records whether it lands on heads or tails. What is the experimental probability that the coin w

zhannawk [14.2K]

Hope this helps you..!

4 0

3 years ago

Read 2 more answers

Please help! I need to finish this before the cutoff!!!!!

katrin2010 [14]

Answer:

im not 100% sure but i think its the last one

D

im not 100% sure though

Step-by-step explanation:

7 0

3 years ago

Solving Rational equations. LCD method. Show work. Image attached.

posledela

$(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12$

So in order to get all the fractions to have a common denominator, we need to multiply $\dfrac k{k-2}$ by $\dfrac{k-6}{k-6}$ , and $\dfrac1{k-6}$ by $\dfrac{k-2}{k-2}$ :

$\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}$

$\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}$

Now,

$\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}$

As long as $k\neq2$ and $k\neq6$ (which we can't have because otherwise $k^2-8k+12=0$ ), we can cancel $k^2-8k+12$ in the denominators on both sides:

$4=(k^2-6k)+(k-2)$

$4=k^2-5k-2$

$0=k^2-5k-6$

We can factorize the right side:

$0=(k-6)(k+1)$

which tells us that $k=6$ and $k=-1$ are solutions.

4 0

4 years ago

All I need is e and f.

vfiekz [6]

E is 1.5 and F is .75

8 0

3 years ago