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2 years ago

Solve 18 and 19 please

Mathematics

1 answer:

riadik2000 [5.3K]2 years ago

5 0

Answer:

Step-by-step explanation:

18.

slope of line y=2x-5 is 2

slope of line ⊥ to y=2x-5 is

m=-1/2

eq. of reqd. line through (10,6) with slope -1/2 is

y-6=-1/2(x-10)

multiply by 2

2y-12=-x+10

2y+x=10+12

2y+x=22

2y=-x+22

y=-1/2x+11

19.

slope of line y=12x+5,m=12

eq. of line through (5,2) with slope 12 is

y-2=12(x-5)

y-2=12x-60

y=12x-60+2

y=12x-58

compare with y=mx+b

b=-58

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Maria and three friends have a thousand and two hundred stock cards if they share the stock of cards equally how many will each

andreev551 [17]

1200 ÷ 4 = 300
Each person will receive 300 stock cards.

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4 years ago

The manufacturer of an energy drink spends $1.20 to make each drink and sells them for two dollars the manufacturer also has fix

dimulka [17.4K]

Let X be the number of energy drinks sold.

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Fixed cost is $8000.

Therefore, cost function is

$C(x)=1.20x+8000$

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Hence, the revenue function is

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6 0

1 year ago

Any help would be appreciated! :)

ikadub [295]

Answer:

-7, then -13, then -19

Step-by-step explanation:

Plug in x values to get y.

y = -2(0) - 7

y = 0 - 7

y = -7

y = -2(3) - 7

y = -6 - 7

y = -13

y = -2(6) - 7

y = -12 - 7

y = -19

Hope this helps. :0)

7 0

3 years ago

Read 2 more answers

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0

4 years ago

Find the lengths of the missing side . Simplify all radicals !!!<br> help mee!!!!!!

larisa86 [58]

Answer:

$e = 13\sqrt{2}$

$f = 13\sqrt{2}$

Step-by-step explanation:

The ∆ given is an isosceles ∆ with a right angle measuring 90°, and two congruent angles measuring 45° each.

Using trigonometric ratio formula, we can find the lengths of the missing side as shown below:

Finding e:

$sin(\theta) = \frac{opp}{hyp}$

$sin(\theta) = sin(45) = \frac{\sqrt{2}}{2}$

hyp = 26

opp = e = ?

Plug in the values into the formula

$\frac{\sqrt{2}}{2} = \frac{e}{26}$

Multiply both sides by 26

$\frac{\sqrt{2}}{2}*26 = \frac{e}{26}*26$

$\frac{\sqrt{2}}{2}*26 = e$

$\frac{\sqrt{2}}{1}*13 = e$

$13\sqrt{2} = e$

$e = 13\sqrt{2}$

Since side e is of the same length with side f, therefore, the length of side f = $13\sqrt{2}$

3 0

3 years ago