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3 years ago

Xy = 6 | 2x – y = 1 Solve each system by substitution

Mathematics

1 answer:

AnnZ [28]3 years ago

7 0

Answer: {x,y} = {2,-4}

[1] x - y = 6

[2] 2x + y = 0

Step-by-step explanation:

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Inequalities<br> Which is the graph of the system x + 3y > -3 and y < 1/2x +1?

OLga [1]

Answer:

the 4th graph.

Step-by-step explanation:

by solving for Y in the first equation, we get the y-intercept of -1 and the second equation has a y-intercept of 1. 4th graph is the only graph with those 2 intercepts

6 0

2 years ago

(25)2×100÷23+[24÷(13−5)]

hram777 [196]

Answer:

100*100/23+(24/8)

100*100/23+3

100*4.37+3

43.7+3

46.7

Step-by-step explanation:

7 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago

Help me<br>in this question<br>

kolbaska11 [484]

hello

Step-by-step explanation:

paano nga Ito???

6 0

3 years ago

If we have 2 bags containing marbles. bag a has 7 green and 6 black marbles. bag b has 8 yellow and 4 red marbles. if we pick ra

Elena L [17]

A. (7/13)·(1/3)=7/39. P=7/39
B. (6/13)·(2/3)=12/39. P=12/39
C. (7/13)·(2/3)=14/39. P=14/39
D. (6/13)·(1/3)=6/39. P=6/39

All the numbers from Bag B were written as thirds cause their probabilities could be simplified to make multiplying easier

8 0

3 years ago