1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alekssr [168]
3 years ago
8

Find the area of the irregular figure

Mathematics
2 answers:
vodomira [7]3 years ago
8 0

Answer:

149 in sq

Step-by-step explanation:

split in to the 3 rectengles

top right 6x(9-6)=18

middle rec 6×(10+6)=96

bottom rec. 5x7=35

total 149

IgorC [24]3 years ago
5 0

Answer:

area = 149 in.²

You might be interested in
Valerie earns $24 per hour. Which expression can be used to show how much money she earns in 7 hours?
GuDViN [60]
B. because you are still multipying they just broke it down (7*24)= 168 , and b also = 168

8 0
3 years ago
Read 2 more answers
Im needing help yall
mario62 [17]

A and B I believe.


It would be A because jump height can be found out by the amount of force you exert to get you into the air minus the gravitational force.


It would also be B because the amount of time you run on the track can be found out by dividing the track length by your speed on the track.


Hope this helps!

5 0
3 years ago
Factor completely 2x2 − 2x − 40
Montano1993 [528]

Answer:

2(x-5)(x+4)

Step-by-step explanation:

2x^2 − 2x − 40

Factor out a 2

2(x^2 -x -20)

What 2 numbers multiply to -20 and add to -1

-5*4 = -20

-5+4 = -1

2(x-5)(x+4)

4 0
3 years ago
Hank earns $12 per hour for babysitting. How much does he earn for 15 hours of babysitting?
Reptile [31]


$180

Multiply 12 by 15.

4 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
Other questions:
  • 135 students attended an assembly at Piper School. The student enrollment at the school is 180. What percentage of the students
    6·1 answer
  • What is 6-1x0+2divided by 3= what<br><br> if someone gets this right I will give them 20 points.
    10·2 answers
  • I have 2 seperate pairs of = sides l. My equal sides are next to each other. My four sides cannot be equal. What am I?
    14·1 answer
  • How do you find the circumference?
    14·2 answers
  • Solve for x in the equation 2x^2+14x+17=-96
    13·1 answer
  • A student is raising money for a local food pantry. A local business agrees to donate an additional 20% of what the student rais
    10·2 answers
  • I need help on this on please
    5·1 answer
  • 10 11/35 divided by 3 4/5
    15·1 answer
  • Which of these systems has one unique solution?
    7·1 answer
  • Evaluate (m4 + 5) – n<br> (m+ + 5) – n3 if m = -2 and n = -4.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!