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3 years ago

Joe solved this problem on a math test. 5/6 - 1/4 = 7/8

Mathematics

1 answer:

Travka [436]3 years ago

5 0

Answer:

This is incorrect, the correct answer is 7/12

Step-by-step explanation:

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Please help will give brainlyisttt

IRINA_888 [86]

Answer:

Either a or c

Step-by-step explanation:

Sorry I'm not entirely sure.

5 0

3 years ago

Read 2 more answers

How would you solve this by elimination?

jekas [21]

To solve this equation by elimination, what you would do is multiply one of the equations by -1, or distribute -1 to each term in the equation, any of the 2 equations. Then align the equations and add them together.

-(X + 3y = 3)
-X - 3y = -3

-X - 3y = -3
X + 6y = 3
__________
3y = 0
y = 0/3 = 0.

Now we can solve for x, by simply plugging the value of y into any of the 2 equations.

X + 6y = 3
X + 6(0) = 3
X + 0 = 3
X = 3.

The solution to your system of equations would be (3,0).

Check this by plugging in the point to the other equation and see if it is true.

X + 3y = 3
(3) + 3(0) = 3
3 + 0 = 3
3 = 3.

Thus it is the solution.

5 0

3 years ago

Write each sequence as a function.

Fiesta28 [93]

Answer: $f(n)=3^{n-1}$

Step-by-step explanation:

Given Recursive formula : $a_{n+1} = 3a_n,$ , $a_1=1$

Then, $a_2=a_{1+1} = 3a_1=3(1)=3$

$a_3=a_{2+1} = 3a_2=3(3)=9$

$a_4=a_{3+1} = 3a_3=3(9)=27$

We can write it as : $f(n)=3^{n-1}$

such that

n $f(n)=3^{n-1}$

1 $f(1)=3^{1-1}=1$

2 $f(2)=3^{2-1}=3^1=3$

3 $f(3)=3^{3-1}=3^2=9$

Hence, the required function: $f(n)=3^{n-1}$

8 0

3 years ago

Read 2 more answers

Barbara Bailey joined a bowling league. She paid a $18.00 entry fee plus $7.50 a week for 16 weeks. She purchased a bowling ball

larisa [96]

If you add them all up it comes out to $339.94

7 0

4 years ago

Read 2 more answers

How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?

irga5000 [103]

The equivalence $n \equiv 5 \pmod{12}$

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5

for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)

we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009

The last k which produces n<5000 is 416

For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.

Answer: 417

6 0

3 years ago