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
Ne4ueva [31]
3 years ago
8

I NEED THE ANSWER ASAP.An architect is planning a triangular pool with the perimeter of 99 feet.If two sides measure 20 and 42,

respectively, what is the length of the missing side which is x?Write an equation to represent the situation, Then solve for x.PLEASE
Mathematics
1 answer:
Alla [95]3 years ago
3 0

Answer:

37 feet

Step-by-step explanation:

99 - (20+42) = 37

You might be interested in
What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.
lapo4ka [179]

Consider the equation x^6+6x^3+5=0.

First, you can use the substitution t=x^3, then x^6=(x^3)^2=t^2 and equation becomes t^2+6t+5=0. This equation is quadratic, so

D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1.

Then you can factor this equation:

(t+5)(t+1)=0.

Use the made substitution again:

(x^3+5)(x^3+1)=0.

You have in each brackets the expression like a^3+b^3 that is equal to (a+b)(a^2-ab+b^2). Thus,

x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)

and the equation is

(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0.

Here x_1=-\sqrt[3]{5} , x_2=-1 and you can sheck whether quadratic trinomials have real roots:

1. D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}.

2. D_2=(-1)^2-4\cdot 1=1-4=-3.

This means that quadratic trinomials don't have real roots.

Answer: x_1=-\sqrt[3]{5} , x_2=-1

If you need complex roots, then

x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2}   ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}.

6 0
4 years ago
What is the value of x ?
OLga [1]

Answer:

x = 20

Step-by-step explanation:

90 = 60 + (x + 10)

90 - 60 = x + 10

30 = x + 10

30 - 10 = x

20 = x

8 0
3 years ago
Read 2 more answers
Use a flowchart to prove if the triangles in each pair are congruent. NO LINKS!!!!​
valentinak56 [21]

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

It isn't clear what is supposed to go in the various blanks. We have elected to identify the corresponding congruent parts, and name the congruent triangles. The postulate supporting the conclusion is also shown.

In most cases, corresponding parts are marked congruent. The exception is the vertical angles in figure 22.

7 0
3 years ago
Read 2 more answers
Am I doing this right I’m trying to help my little brother with his math homework but I don’t remember how to do this and he wan
NeX [460]

Answer:

The least common multiple is 30. Hope that helps!!

Step-by-step explanation:

5 0
3 years ago
Point P is located -13, and point Q is located at 8.
marshall27 [118]
2

Hope this helps

Mark brainliest please
5 0
2 years ago
Other questions:
  • Ms. Luna is waterproofing the top of a rectangular wood deck. The width of the deck is 3 m less that the length. The length is 8
    15·1 answer
  • Find exact length of the third side
    11·1 answer
  • How many different arrangements of the letters in the word PARALLEL are there ?Type a numerical answer in the space
    5·1 answer
  • over a period of two weeks, Todd went from bench pressing 190 pounds to 220 pounds. Find the increase in weight that Todd can be
    7·2 answers
  • Ok what is 4 1/8 - 3/4
    11·2 answers
  • What is the common ratio for the following geometric sequence? -500, 100 , -20, 4,....
    11·1 answer
  • An electrician collected $1265.23 for a job that included $840 labor, which was not taxed, and the rest for parts, which were ta
    12·1 answer
  • I need help with this question answer ASAP
    14·1 answer
  • Please help for brainliest !!
    8·1 answer
  • HellpppNeed help please urgent
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!