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3 years ago

2x + 3y = 23 -1(2x + y = 13

Mathematics

2 answers:

notsponge [240]3 years ago

6 0

Answer:

y= 9

x = -2

Step-by-step explanation:

Y= 9, X = -2 I use the simultaneous equation to solve it.

vivado [14]3 years ago

3 0

Answer:

2x +3y= 5xy

-1(2x+y= 2xy+-1 = -3xy

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hich uses the high-low method to analyze cost behavior, has determined that machine hours best predict the company's total ut

Crank

Answer:

1)- Variable utilities cost per machine hour = 1.6 per machine hour

2)- Fixed cost = 1740

3)-Total cost on 1220 Machine hour will be

= 3692

Step-by-step explanation:

1) CALCULATE VARIABLE UTILITIES COST PER MACHINE HOUR :

Variable utilities cost per machine hour = Change in cost/high machine hour-low machine hour

=4076-3388/1460-1030

Variable utilities cost per machine hour = 1.6 per machine hour

2) Fixed cost = Total cost-variable cost

= 3388-(1030*1.6)

Fixed cost = 1740

3) Total cost on 1220 Machine hour will be (1220*1.6+1740) = 3692

3 0

3 years ago

Solve –3x = 36 for x. A. x = 9 B. x = 12 C. x = –12 D. x = –9

lawyer [7]

Answer: C

Step-by-step explanation:

-3x=36

Divide -3 by both sides

x=-12

8 0

3 years ago

9th grade math please help

Leni [432]

Answer:

Only:

B. x=7

D. x=0

Step-by-step explanation:

Substitute the answers in the formula and only 7 and 0 will work.

6 0

3 years ago

Alex purchased a seven-year $600 bond at par value with a 5% coupon. What is the total value of the coupons?

anyanavicka [17]

Answer

300 dollars

because u do 600x.5

4 0

3 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$

from (1)(2) => proven

5 0

3 years ago