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Alex Ar [27]
2 years ago
11

How did brianly choose the best answer

Mathematics
1 answer:
ladessa [460]2 years ago
5 0

Answer:

➢ Unfortunately, Brainly appears more to be a place to get quick answers so that students can finish homework faster. The answers, even if they're correct, often offer little to no elaboration or depth on the topic, and there's no guarantee that given answers will be correct.

#CarryOnLearning

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Worth 48 points!!!! plz help me.
Mrac [35]

Answer:

see below

Step-by-step explanation:

1.  The inverse means take the not of both sides

If it is Saturday, then there is no school.

 Inverse     If it is not Saturday, then there is school.

The converse flips the then and the if

If there is no school, then it is Saturday.

2.) The first given part is the if, what will happen is the then part

If you are a freshman, then you are required to attend orientation.

3.)

Given: 3x+1=-14 Prove: x=-5

Statements    Reasons

1. 3x+1=-14     Given

2. 3x=-15          Subtraction Property of equality  ( since we subtract 1 from                         each side)

3. X=-5          Division Property  ( since we divide each side by 3)

3 0
3 years ago
Read 2 more answers
52.5 + s I need help please
sleet_krkn [62]

Answer:

try 00.1

Step-by-step explanation:

6 0
2 years ago
Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!
il63 [147K]

Answer:

x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

Step-by-step explanation:

w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

6 0
3 years ago
Find all solutions to
BARSIC [14]

Answer:

x= 0 , \frac{1}{14} , \frac{-1}{12}

Step-by-step explanation:

Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x}. →→→ (1)

Now, by cubing the equation on both sides, we get

( \sqrt[3]{15x-1} + \sqrt[3]{13x+1} )³ = (4\sqrt[3]{x})³

⇒ (15x-1) + (13x+1) + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1}) = 64 x.

⇒ 28x + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (4\sqrt[3]{x}) = 64x.        

(since from (1),  \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x})

⇒ 12× \sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{x})= 36x.

⇒ 3x = \sqrt[3]{(15x-1)(13+1)(x)}.

Now, once again cubing on both sides, we get

(3x)³ = (\sqrt[3]{(15x-1)(13+1)(x)})³.

⇒ 27x³ = (15x-1)(13x+1)(x).

⇒ 27x³ = 195x³ + 2x² - x

⇒ 168x³ + 2x² - x = 0

⇒ x(168x² + 2x -1) = 0

⇒ by, solving the equation we get ,

x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}

therefore, solution is x= 0 , \frac{1}{14} , \frac{-1}{12}

7 0
3 years ago
Evaluate the expression.<br><br> (132−122)÷5<br> The value of the expression___ is
ludmilkaskok [199]
The answer is 107.6 pemdas order of operations
3 0
3 years ago
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