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Natasha2012 [34]
3 years ago
15

Find roots of 4x^2-2x+9

Mathematics
1 answer:
Korolek [52]3 years ago
4 0

discriminant= b^2-4ac= (-2)^2-4(4)(9)= -140<0

since discriminant <0, graph is always positive, x has no real roots

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The sum of two numbers is 30. the first number is 1/5 of the second number. what are the numbers?
bonufazy [111]
Same question
x+y=30
x is 1/5 of y
x=y/5
subsitute
y/5+y=30
multily both sidse by 5
y+5y=150
add
6y=150
divide both sides by 6
y=25
subsitute
x=y/5
x=25/5
x=5
the numbers are 5 and 25

6 0
3 years ago
Find all solutions in the interval (0,2pi) you should get kpi and pi/4 but how do you get there
Ket [755]

Add sin(x) to both sides

tan(x)sin(x)=sin(x)

Divide sin(x) from both sides

tan(x)=1

now you just have to figure out where tan(x)=1. you can figure this out from there by looking at the unit circle.

4 0
3 years ago
What is <br> 10x + 300 = 800<br> Pls
kupik [55]

Answer:

x = 50

Step-by-step explanation:

10x + 300 = 800

subtract 300 from both sides

10x = 500

divide each side by 10

x = 50

8 0
3 years ago
Read 2 more answers
Trouble finding arclength calc 2
kiruha [24]

Answer:

S\approx1.1953

Step-by-step explanation:

So we have the function:

y=3-x^2

And we want to find the arc-length from:

0\leq x\leq \sqrt3/2

By differentiating and substituting into the arc-length formula, we will acquire:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx

To evaluate, we can use trigonometric substitution. First, notice that:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx

Let's let y=2x. So:

y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx

We also need to rewrite our bounds. So:

y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0

So, substitute. Our integral is now:

\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Let's multiply both sides by 2. So, our length S is:

\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

y=a\tan(\theta)

Substitute 1 for a. So:

y=\tan(\theta)

Differentiate:

y=\sec^2(\theta)\, d\theta

Of course, we also need to change our bounds. So:

\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0

Substitute:

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta

The expression within the square root is equivalent to (Pythagorean Identity):

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta

Simplify:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta

And:

dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)

Integration by parts:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)

Distribute:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)

Now, let's make the single integral into two integrals. So:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Distribute the negative:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Divide the second and third equation by 2. So: \displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))

Therefore, our arc length will be equivalent to:

\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Divide both sides by 2:

\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Evaluate:

S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))

Evaluate:

S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))

Simplify:

S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}

Use a calculator:

S\approx1.1953

And we're done!

7 0
3 years ago
An airplane descends during the last hour of it's flight to prepare for landing. It's altitude changes at an average of -0.15 km
Natalija [7]

Answer:

-.15 km/ minute * 60 minutes

-9 km

Step-by-step explanation:

The rate is -.15 km per minute

We have 60 minutes

distance = rate  times time

change in elevation is the same as the distance change

change in elevation = -.15 km/ minute * 60

   change in elevation =-9 km

8 0
3 years ago
Read 2 more answers
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