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3 years ago

2.5y=5 what is the answer to the equation?

Mathematics

2 answers:

postnew [5]3 years ago

7 0

Answer:2

Step-by-step explanation:

2.5(2)

DedPeter [7]3 years ago

5 0

Answer:

y=.5

Step-by-step explanation:

Set 2.5 as 25 and weal come back to the deci. 25y=5 is the same as 25/5 there for you get the answer 5 but now you add the decimal getting you to .5 . welcome

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Anna buys 12 pens and 20 pencils for a total of $16. One pen costs $0.40 more than one pencil. What would 1 pen and 1 pencil cos

nevsk [136]

Answer:

Cost of pencil is $0.35 and cost of pen is 0.75

Step-by-step explanation:

The cost of one pencil and one pen is required.

The cost of a pencil is $0.35 and a pen is $0.75.

The number of pens is 12.

The number of pencils is 20.

Let the cost of one pencil be

A pen costs $0.40 dollar more than pen.

So, cost of a pen is

Total cost of all the pens and pencils is $16.

Cost of pencil is $0.35 and cost of pen is 0.75

U 2 can help me by marking as brainliest.......

5 0

2 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

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Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.