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3 years ago

Whats the vertex of the graph y=-x^2

1 answer:

6 0

(0,0) you could use x=-b/2a but since your b term is nonexistant the whole term goes to 0 for the x value. Plug x=0 into the original equation which gives you another 0. ezpz

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Does anybody know how to do this

Nitella [24]

Answer: B

Reason: Plugged it in in my Desmos app, it's free on the appstore and helps you solve for graphs !

7 0

3 years ago

Dylan owes half as much money as he used to owe. If he used to owe $28, which of the following expressions would reflect how muc

zubka84 [21]

Answer:

Step-by-step explanation:

28 ÷ 2 equals 14 and 14 is half of 28

4 0

3 years ago

Read 2 more answers

F(x)=3.7-2x

Luda [366]

Answer:

C) h(x)=-1.3-1.75x

Step-by-step explanation:

f(x)=3.7-2x

g(x) = 0.25x-5

f(x) + g(x) =3.7-2x + 0.25x-5

Combine like terms

= -1.75x - 1.3

5 0

3 years ago

Please help me with this question step by step on how to solve it. It takes the earth 24 h to complete a full rotation. If it wo

pav-90 [236]

In order to find the number of hours it takes to do the full rotation we separate into pieces the days, hours, and minutes and convert each of them separately.

using the conversion factor from days to hours we get that

$1\text{day}=24\text{hours}$

then we get that

$38\text{days}\cdot\frac{24\text{hours}}{1day}=912hours$

hours does not need a conversion factor, meaning that

$10\text{hours}=10\text{hours}$

continue by converting the minutes using the following conversion factor

$1\text{hour}=60\min$

then,

$30\min \cdot\frac{1\text{hour}}{60\min }=0.5\text{hours}$

To complete add all the results together

$\begin{gathered} 912+10+0.5 \\ 922.5\text{hours} \end{gathered}$

It takes mercury 922.5 hours to complete a full rotation.

3 0

1 year ago

Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif

xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0

3 years ago