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patriot [66]
3 years ago
5

Which is a correct way of graphing the line y = –x – 2?

Mathematics
1 answer:
katen-ka-za [31]3 years ago
4 0
Y = -x - 2.....in y = mx + b form, the m represents the slope and the b represents the y int. (and if u didnt already know, -x is the same as -1x)

so in ur equation, the slope is -1 and the y intercept is -2....or (0,-2)

so u would plot (0,-2) and use the slope of -1 to plot another point. Draw a line through the points.
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IF U CAN READ WHAT IT SAYS uu JUST MIGHT GET A BRAINLIEST
Nutka1998 [239]

Answer:

hmmm...

Step-by-step explanation:

This looks like...

"jnxxmppoihuueitsdehbxexdnjussheyetyyuuhggfffwspiritualdjjdmsmmsmyra."

YES!

(lol)

4 0
3 years ago
A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I co
zubka84 [21]

Answer:

a) i The company should buy 40 gallons from dairy I and 60 gallons from dairy

ii) What is the maximum amount of​ butterfat? The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b.The excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

Step-by-step explanation:

a. How much milk from each supplier should the company buy to get at most 100 gallons of milk with the maximum amount of​ butterfat?

From the question, we are told that:

Milk from dairy I costs ​$2.40 per​ gallon, Milk from dairy II costs ​$0.80 per gallon.

Let's represent:

Number of gallons of Milk from dairy I = x

Number of gallons of Milk from dairy II = y

At most ​$144 is available for purchasing milk.

$2.40 × x + $0.80 × y = 144

2.40x + 0.80y = 144........ Equation 1

x + y = 100....... Equation 2

x = 100 - y

2.40(100 - y) + 0.80y = 144

240 - 2.4y + 0.80y = 144

-1.60y = 144 - 240

-1.6y = -96

y = -96/-1.6

y = 60

From Equation 2

x + y = 100....... Equation 2

x + 60 = 100

x = 100 - 60

x = 40

Therefore, since number of gallons of Milk from dairy I = x and number of gallons of Milk from dairy II = y

The company should buy 40 gallons from dairy I and 60 gallons from dairy

II. What is the maximum amount of​ butterfat?

From the question

Dairy I can supply at most 50 gallons averaging 3.9​% ​butterfat,

50 gallons = 3.9% butterfat

40 gallons =

Cross Multiply

= 40 × 3.9/50

= 3.12%

Dairy II can supply at most 90 gallons averaging 2.9​% butterfat.

90 gallons of milk = 2.9% butter fat

60 gallons of milk =

Cross Multiply

= 60 × 2.9%/90

=1.9333333333%

≈ 1.93%

The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b. The solution from part a leaves both dairy I and dairy II with excess capacity. Calculate the amount of additional milk each dairy could produce.

Excess capacity of Diary I =

50 gallons - 40 gallons = 10 gallons

Excess capacity of Diary II =

90 gallons - 60 gallons = 30 gallons

Therefore, the excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

3 0
3 years ago
Original cost: $7.50; markup %:50%
vagabundo [1.1K]
You would pay $3.75
7.50÷2
6 0
3 years ago
Read 2 more answers
1/2x - 7 = 1/3(x-12) what is x?
sashaice [31]

Answer:

x = 18

Step-by-step explanation:

took test!!

7 0
3 years ago
Read 2 more answers
the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i
nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

A = \frac{1}{2}\cdot b\cdot h (Eq. 1)

Where:

A - Area of the triangle, measured in square centimeters.

b - Base of the triangle, measured in centimeters.

h - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt} (Eq. 2)

Where:

\frac{dA}{dt} - Rate of change of area in time, measured in square centimeters per minute.

\frac{db}{dt} - Rate of change of base in time, measured in centimeters per minute.

\frac{dh}{dt} - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

\frac{1}{2}\cdot\frac{db}{dt}\cdot h =  \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}

\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt} (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

b = \frac{2\cdot A}{h}

If we know that A = 130\,cm^{2}, h = 15\,cm, \frac{dh}{dt} = 2.5\,\frac{cm}{min} and \frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}, the rate of change of the base of the triangle in time is:

b = \frac{2\cdot (130\,cm^{2})}{15\,cm}

b = 17.333\,cm

\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)

\frac{db}{dt} = -2.262\,\frac{cm}{min}

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

6 0
3 years ago
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