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3 years ago

IF ANYONE HELP ME OUT WITH THESE PROBLEMS ILL GIVE EXTRA POINTS AND BRAINLEST

Mathematics

2 answers:

fiasKO [112]3 years ago

6 0

Answer:

it's acute

${65 }^{2} + {72 }^{2} > {96}^{2}$

4225+5184>9216

9409>9216

so it's acute

if a2+b2>c2 ---acute

a2+b2=c2 ---- right angle

a2+b2<c2 --obtuse

Dennis_Churaev [7]3 years ago

5 0

Answer:

it's obtuse. reason being cause to be acute it has to be under 90 and to be right it has to be at 90

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Which equation is equivalent to StartRoot x squared + 81 EndRoot = x + 10?

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The given equation is equivalent to Option D .

The mission option is

A. x+10= x+10

B. x+9 =x² + 20x + 100

C. x² +81 = x² + 100

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<h3>What is an Equation ?</h3>

An equation is a mathematical statement formed when two algebraic expression is equated by an equal sign .

An equation is given in the question

$\rm \sqrt{x^2 + 81} = x +10$

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2 years ago

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Answer:

Step-by-step explanation:

First both the rational numbers should have same denominators. So, find least common denominator

Least common denominator is 10

$\frac{-1}{5}=\frac{-1*2}{5*2}=\frac{-2}{10}\\\\\\\frac{1}{2}=\frac{1*5}{2*5}=\frac{5}{10}\\\\\\$

Now multiply the numerator and denominators of the both the numbers by 10.

$\frac{-2*10}{10*10}= \frac{-20}{100} \ and \ \frac{5*10}{10*10}=\frac{50}{100}\\\\\\Now\ 10 \ rational \ numbers \ between\frac{-20}{10} \ and \ \frac{50}{100} \ is \\\\\\\frac{-19}{100},\frac{-18}{100},\frac{-17}{100},\frac{-16}{100}.\frac{-15}{100},\frac{-14}{100},\frac{-13}{100},\frac{-12}{100},\frac{-11}{100};\frac{-10}{100}$

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3 years ago

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1 year ago

Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)

Svet_ta [14]

<h2>Answer:</h2>

$\boxed{x=1, \y=-1, \ z=2}$

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

$\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}$

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

$\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}$

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

$\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}$

Step 4: solve for z, then for y, then for x:

$\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}$

$-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}$

By substituting $y=-1 \ and \ z=2$ into the first equation, we get the $x$ . So:

$x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}$

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4 years ago

Read 2 more answers

IF ANYONE HELP ME OUT WITH THESE PROBLEMS ILL GIVE EXTRA POINTS AND BRAINLEST​

IF ANYONE HELP ME OUT WITH THESE PROBLEMS ILL GIVE EXTRA POINTS AND BRAINLEST