All categories

2 years ago

15 points!!! Write an equation of the parabola shown.

Mathematics

1 answer:

Serjik [45]2 years ago

8 0

Answer:

y=1/2x-4

Step-by-step explanation:

find the gradient and sub in to find y intercept.give me brainliest pls this took forever to do,

You might be interested in

Find three consecutive integers such as two times the first integer, plus three times the second integer, minus the third intege

NikAS [45]

The three consecutive integers are 5, 6, and 7.

What are consecutive integers?

Whole numbers that follow one another without a gap are known as consecutive integers. A few examples of consecutive integers are 15, 16, and 17, 1,2 and 3, and so on.

Finding the Consecutive Integers

Let the three consecutive integers be a, a + 1, and a + 2.

These three consecutive integers are to be such that two times the first integer when added to three times the second integer minus the third integer, it is equal to 21.

⇒ 2a + 3(a+1) - (a+2) = 21

2a + 3a + 3 - a -2 = 21

5a - a + 3 - 2 = 21

4a + 1 = 21

4a = 20

a = 20/4

a = 5

∴ a+1 = 5+1

a+1 = 6

And, a+2 = 5 + 2

a +2 =7

Hence, the three consecutive integers come out to be 5, 6, and 7

Learn more about consecutive integers here:

brainly.com/question/1767889

#SPJ4

8 0

1 year ago

Are the graphs of y= 1/3x-4 and y-2=-3(x-4) parallel, perpendicular, or neither

melamori03 [73]

Answer:

ghtrh

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What's 5/6 divided by 2

natali 33 [55]

5/6÷2/1
=5/6×1/2
=5/12 i hope this helps

7 0

2 years ago

Which table shows a proportional relationship between x and y? A. x y 2 4 3 6 4 9 B. x y 3 4 9 16 15 20 C. x y 4 12 5 15 6 18 D.

Sliva [168]

Answer:

Step-by-step explanation:

The relationship is proportion if y/x is constant.

So it is C because 12/4 = 15/5 = 18/6 = 3.

7 0

2 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago