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2 years ago

2. 4x + 5y = -30 Slope-standard form

Mathematics

1 answer:

Lerok [7]2 years ago

7 0

Y=mx+c is the slope standard form
5y=-30-4x
5y=-4x-30
Y=-4/5x-6

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The sum of two numbers is 52. If one number is 76 times the second number, then find the largest number out of the two.

butalik [34]

Answer:

72 +1 , If u = A 8 1 - 2 ( ( + + +1 )– flog ( 3 + 3 / 3 ) ? v1m * + n + 13 ) 4 1 ; then 2 If c ... of any number of such MATHEMATICS . triangles ; and ( 3 ) find the mean value of ... 72 = 62 + 69 , 73 = 82 + 39 , 75 = 72 + 52 . the in - centre and centroid . ... 4n + 1 , or 4n + 2 ,

Step-by-step explanation:

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2 years ago

Bananas are 5 for 79 cents . Isaiah bought 7 bananas. How many did Isaiah spend?

Margarita [4]

C it’s always c. When in doubt always chose c

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3 years ago

Which models show 40%? Check all that apply. A model with 4 shaded sections and 6 unshaded sections. A model with 40 shaded sect

Amanda [17]

Answer:

A model with 4 shaded sections and 6 unshaded sections.

A model with 40 shaded sections and 60 unshaded sections.

Step-by-step explanation:

Required:

Which models shows 40%

(a) 4 shaded and 6 unshaded

First, we calculate the total sections

$Total = 4 + 6$

$Total = 10$

The percentage of the shaded section is:

$Shaded = \frac{4}{10} * 100\%$

$Shaded = 0.4 * 100\%$

$Shaded = 40\%$

(b) 40 shaded and 60 unshaded

First, we calculate the total sections

$Total = 40 + 60$

$Total = 100$

The percentage of the shaded section is:

$Shaded = \frac{40}{100} * 100\%$

$Shaded = 0.4 * 100\%$

$Shaded = 40\%$

(c) 2 shaded and 8 unshaded

First, we calculate the total sections

$Total = 2 + 8$

$Total = 10$

The percentage of the shaded section is:

$Shaded = \frac{2}{10} * 100\%$

$Shaded = 0.2 * 100\%$

$Shaded = 20\%$

<em>Hence, (a) and (b) shows 40%</em>

8 0

2 years ago

Read 2 more answers

there has to be at least one operating path. Once a component fails on a path, that path is no longer operating. The reliability

Eva8 [605]

Answer:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

Step-by-step explanation:

We can assume that the system work in series

If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

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3 years ago

If a-1/a = 8 find the value of (a+1/a) square

Strike441 [17]

Answer:

(1) a+1/a=8a+2/8a+1

(2) a²-1/a²=(a+1)(a-1))a²

(8a+2)(8a)/(8a+1)²

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2 years ago