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
Vera_Pavlovna [14]
3 years ago
11

a right triangle has leg lengths of 1 foot 6 inches and 2 feet. find the hypotenuse length and the perimeter in mixed units of f

eet and inches
Mathematics
2 answers:
MariettaO [177]3 years ago
8 0
The hypotenuse is 2 feet or 24 inches cuz it is the larget. the perimeter is 3.5 feet or 42 inches
zhannawk [14.2K]3 years ago
4 0
1ft 6 in = 18 in
2ft = 24 in
Hyp length = sq root (18^2 +24^2) =30in = 2ft 6in
Perimeter = 1ft 6in + 2 ft + 2ft 6in = 6ft
You might be interested in
Quadrilateral ABCD has coordinates A (3,1) B (4,4) C (7,5) D (6,2). Quadrilateral ABCD is a ?
ASHA 777 [7]
The Quadrilateral ABCD with coordinates (3, 1), (4, 4), (7, 5), (6, 2) is a rhombus because its length and width are both square root of 10 units and adjacent sides are not perpendicular. This can be seen by plotting the points you can either plot it or by using programs.

I hope this helps, God bless, and have a great day!
Brainliest is always appreciated :)
3 0
3 years ago
Read 2 more answers
Can someone please answer all the questions with explanations?
aalyn [17]

Answer:

I don't know sorry can't help you

6 0
2 years ago
Read 2 more answers
Rich is building a travel crate for his dog, Thomas, a beagle-mix who is about 28 inches long, 11 inches wide, and 24 inches tal
Dmitry [639]

Answer:

27600

Step-by-step explanation:

28 long +12 -->40

11 wide +12 --->23

24 tall +6--->30

v=40×23×30=27600 cubic inches

4 0
2 years ago
Read 2 more answers
Can anyone help me ???
lana [24]
The slope would be positive in this case
8 0
3 years ago
Read 2 more answers
How do you find the limit?
coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}

We can cancel x-5.

\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}

Then directly substitute x = 5 in.

\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}

<u>Differentiation</u> (Property of Addition/Subtraction)

\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}

Hence from the expressions,

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}

<u>Differential</u> (Constant)

\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \  is \ \ a \ \ constant.)}}

Therefore,

\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}

Now we can substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}

Thus, the limit value is 2/5 same as the first method.

Notes:

  • If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.
8 0
3 years ago
Other questions:
  • Need help quickly please
    8·1 answer
  • Why is graphing not always the best option for solving systems?
    10·2 answers
  • 0.48 is ____% of 1.6. Do not round you answer.
    15·2 answers
  • Express the statement "A rectangular field with an area of 120m' has it length 12mn longer than the width" as quadratic equation
    11·1 answer
  • Please solve 15x+1=10+2x
    6·2 answers
  • In ΔCAB, point E is the midpoint of segment AC and point D is the midpoint of segment BC. If the measure of segment AB is 8 unit
    6·2 answers
  • Please help! Will mark brainliest
    5·2 answers
  • Please answer and show the work! Will reward 100 points
    6·2 answers
  • How many dots are there at 100 minutes?
    13·2 answers
  • The screen shots are attatches please help i will give brainliest. inches to miles
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!