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3 years ago

What do I do next???

Mathematics

2 answers:

Novay_Z [31]3 years ago

5 0

Answer: You have to get rid of the variables (the x's) so subtract 2x from 2x and 11x then go from there.

Step-by-step explanation:

Elina [12.6K]3 years ago

4 0

Answer:

3 and 6

Step-by-step explanation:

11x - 53 = x^2 + 2x - 35

Subtract 11x - 53 on both sides,

0 = x^2 + 2x - 35 - ( 11x - 53 )

= x^2 + 2x - 35 - 11x + 53

= x^2 + 2x - 11x + 53 - 35

0 = x^2 - 9x + 18

x^2 - 9x + 18 = 0

Now this equation need to factorized.

x^2 - 9x + 18 = 0

x^2 - 3x - 6x + 18 = 0

x ( x - 3 ) - 6 ( x - 3 ) = 0

( x - 3 ) ( x - 6 ) = 0

x - 3 = 0

x = 3

x - 6 = 0

x = 6

The solutions to this system are 3 and 6.

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Solve for X

tigry1 [53]

Answer:

$x = \pm e^{-1}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
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Algebra II

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Step-by-step explanation:

Step 1: Define

$\displaystyle ln(x^2) = -2$

Step 2: Solve for x

Raise both sides by e: $e^{\displaystyle ln(x^2)} = e^{-2}$
Simplify equation: $x^2 = e^{-2}$
Square root both sides: $x = \pm e^{-1}$

7 0

3 years ago

Read 2 more answers

What is the value of x with clear working out please?

nirvana33 [79]

Answer:

Answer:  x is 46° 

Step-by-step explanation:

• Let's first find Angle ACB

${ \rm{ \angle ACB + 23 \degree + 90 \degree = 180 \degree}} \\ \\ { \rm{ \angle ACB = (180 - 90 - 23) \degree}} \\ \\ { \underline{ \rm{ \: \angle ACB = 67 \degree}}}$

• From alternative angles, x = Angle BAC.

• Since AB = AC, then Angle ABC = Angle ACB

${ \rm{x + \angle ABC + \angle ACB = 180 \degree}} \\ \\ { \rm{x + 67 \degree + 67 \degree = 180 \degree}} \\ \\ { \rm{x + 134 \degree = 180 \degree}} \\ \\ { \boxed{ \rm{x = 46 \degree}}}$

5 0

3 years ago

Read 2 more answers

Which trigonometric identity can be used to prove the given statement? tan0 cos0 = sin0

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

Free trigonometric identity calculator - verify trigonometric identities step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy.

4 0

2 years ago

Cosx+1/sin^3x=cscx/1-cosx

ANTONII [103]

I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).


If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).


Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.


I hope this helps!

8 0

3 years ago

PLZ HELP ILL MARK U BRAINLIEST!!

pychu [463]

Answer:

Step-by-step explanation:

The two horizontal line segments are the parallel sides of the trapezoid.

∠1 and ∠2 are corresponding angles of a transversal across parallel lines, so

∠1 is congruent with ∠2

5 0

3 years ago

What do I do next???​

What do I do next???