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
Georgia [21]
3 years ago
7

NEED HELP, ASAP! !!! 2 QUESTIONS.

Mathematics
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

1.we know the perimeter of the triangle=x-3+x+6+x=3x+3

so,answer is (b)3x+3

2.area of the triangle=1/2*b*h

=1/2*x+2*2x+6=x+2*2x+6/2=2x^2+6x+4x+12/2=2x^2+10x+12/2=2(x^2+5x+6)/2=x^2+5x+6

so,answer is(b)x^2+5x+6

You might be interested in
Find the volume of the solid generated when R​ (shaded region) is revolved about the given line. x=6−3sec y​, x=6​, y= π 3​, and
Dmitrij [34]

Answer:

V=9\pi\sqrt{3}

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

V=\int\limits^b_a {\pi r^{2}} \, dy

where:

r=6-(6-3 sec(y))

r=3 sec(y)

a=0

b=\frac{\pi}{3}

so the volume becomes:

V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy

This can be simplified to:

V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy

and the integral can be rewritten like this:

V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy

which is a standard integral so we solve it to:

V=9\pi[tan y]\limits^\frac{\pi}{3}_0

so we get:

V=9\pi[tan \frac{\pi}{3} - tan 0]

which yields:

V=9\pi\sqrt{3}]

6 0
3 years ago
PLEASE HELP I WILL REWARD BRANLIEST AND 99 POINTS!!!!
ch4aika [34]

Answer:

Step 3, because the solution should also include all the values for x between the two given values

Step-by-step explanation:

step 1

we have

y_1=\left|x-4\right|

y2=1

step 2

The graph in the given problem

step 3

Identify the solution when y1\leq y2

so

\left|x-4\right|\le1

using a graphing tool

The solution is the interval  -----> [3,5]

see the attached figure

All real numbers greater than or equal to 3 and less than or equal to 5

therefore

The first step in which the student made an error is step 3

6 0
4 years ago
Read 2 more answers
The model represents an equation.
madreJ [45]

Answer:

9/5

Step-by-step explanation:

solve the original equation for x

-4× + 3 = 1x +(-6)

-1x

-5x + 3 = -6

-3

-5x = -9

/-5

x = 9/5

CHECK

-4(9/5) + 3 = 1(9/5) +(-6)

-21/5 = -21/5

8 0
3 years ago
What is the value of x? Round to the nearest tenth. Please Need Help Badly.
Maksim231197 [3]

Answer: IDK hope you get a good grade :0 SORRY

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Create an equation that you can use in everyday life. Then solve it.
galina1969 [7]
If you were to plan a party, you could use a linear equation. for example, the room you are renting is $500. The price for food per person is $10

5 0
3 years ago
Other questions:
  • a school has 520 students. dan surveys a random sample of 50 students and finds that 32 have pet cats. how many students are lik
    11·2 answers
  • a group of 4 people share 1/2 pound of almonds. how many pounds of almonds does each person get choose an equation you can use t
    6·2 answers
  • A cell phone provider charges $30 each month to have a cell on their plan, but their talk and text are not free. It costs $0.05
    10·2 answers
  • What is the length of the diagonal, d, of the rectangular prism shown below?
    10·2 answers
  • Identify the terms, like terms, coefficients, and constants in the expression 5a+2+7+6a
    8·2 answers
  • Joe has a few choices for
    14·1 answer
  • 40 points Choose the definition for the function
    13·1 answer
  • Martha (from problem 9) has some expenses each week. Each week she buys $9.72 worth of
    9·2 answers
  • 3(1 - 2x) &gt; 3 - 6x<br> help-
    13·1 answer
  • Check cashing companies
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!