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3 years ago

Solve the equation: 12=19-√3x-11

Mathematics

2 answers:

blondinia [14]3 years ago

5 0

Step 1 :Isolate the square root on the left hand side :

Original equation
 12 = 19-√3x-11

 Isolate
 √3x-11 = -12+19

 Tidy up
 √3x-11 = 7

Step 2 :Eliminate the radical on the left hand side :

Raise both sides to the second power
 (√3x-11)2 = (7)2

 After squaring
 3x-11 = 49

Step 3 :Solve the linear equation :

Rearranged equation
 3x -60 = 0

 Add 60 to both sides
 3x = 60

 Divide both sides by 3
 A possible solution is :
 x = 20

Step 4 :Check that the solution is correct :

Original equation, root isolated

√3x-11 = 7

 Plug in 20 for x
 √3•(20)-11 = 7
 Simplify
 √49 = 7
 Solution is:
 x = 20

One solution was found : x = 20

Alekssandra [29.7K]3 years ago

4 0

The answer is x=20. You isolate the square root. Eliminate the radical on the left handside. And then solve it!

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Given f(x) = 3x + 1, solve for x when f(x) = 7.

irakobra [83]

Answer:

f(7) = 22 or x =2

Step-by-step explanation:

f(x) = 3x + 1
f(7) = 3(7) + 1
f(7) = 21 + 1
f(7) = 22

Or if the equation meant it like this:

f(x) = 3x + 1
7 = 3x + 1
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tino4ka555 [31]

Step-by-step explanation:

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I need to know the improper fractions answers for:

zmey [24]

Answer:

Part 1) $x=\frac{15}{2}\ units$

Part 2) $z=\frac{15\sqrt{3}}{2}\ units$

Part 3) $y= \frac{15\sqrt{3}}{4}\ units$

Part 4) $b= \frac{45}{4}\ units$

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

$cos(60^o)=\frac{x}{15}$ ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

$cos(60^o)=\frac{1}{2}$

substitute

$\frac{1}{2}=\frac{x}{15}$

solve for x

$x=\frac{15}{2}\ units$ ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

$15^2=x^2+z^2$

substitute the value of x

$15^2=(\frac{15}{2})^2+z^2$

solve for z

$z^2=15^2-(\frac{15}{2})^2$

$z^2=225-\frac{225}{4}$

$z^2=\frac{675}{4}$

$z=\frac{\sqrt{675}}{2}\ units$

simplify

$z=\frac{15\sqrt{3}}{2}\ units$

step 3

Find the value of y

In the right triangle of the right

$sin(30^o)=\frac{y}{z}$ ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$sin(30^o)=\frac{1}{2}$

$\frac{1}{2}=y:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{1}{2}= \frac{2y}{15\sqrt{3}}$

$y= \frac{15\sqrt{3}}{4}\ units$

step 4

Find the value of b

In the right triangle of the right

$cos(30^o)=\frac{b}{z}$ ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}$

$b= \frac{45}{4}\ units$

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solmaris [256]

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