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2 years ago

Solve for .....x 2x+8=4

Mathematics

2 answers:

Lynna [10]2 years ago

5 0

Answer:

x= -2

Step-by-step explanation:

isolate x

2x+8=4 subtract 8 on both sides

2x=-4 divide 2 on both sides

x= -2

Naya [18.7K]2 years ago

4 0

Answer:

2x=-4

x=-2

Step-by-step explanation:

You might be interested in

Find the value of x.<br> х= 2<br> х= 3<br> х= 33<br> х= 52

nlexa [21]

Answer:

x = 3

Step-by-step explanation:

The external angle of a triangle is equal to the sum of the 2 opposite interior angles

∠ QRT is an external angle, thus

∠ QRT = ∠ RTS + ∠ RST, substitute values

45x = 25x + 57 + x

45x = 26x + 57 ( subtract 26x from both sides )

19x = 57 ( divide both sides by 19 )

x = 3

6 0

2 years ago

Help plz and thank you.I don't know what am I suppose to do on this problems?!

lesya [120]

D= domain = farthest left and right
Domain = (3, infinite)
R = range = bottom to top
Range = (2,2)

3 0

2 years ago

Can someone give me the answers and step by step instructions please??

professor190 [17]

Answer:

$-1,4,-7,10,...$ neither

$192,24,3,\frac{3}{8},...$ geometric progression

$-25,-18,-11,-4,...$ arithmetic progression

Step-by-step explanation:

Given:

sequences: $-1,4,-7,10,...$

$192,24,3,\frac{3}{8},...$

$-25,-18,-11,-4,...$

To find: which of the given sequence forms arithmetic progression, geometric progression or neither of them

Solution:

A sequence forms an arithmetic progression if difference between terms remain same.

A sequence forms a geometric progression if ratio of the consecutive terms is same.

For $-1,4,-7,10,...$ :

$4-(-1)=5\\-7-4=-11\\10-(-7)=17\\So,\,\,4-(-1)\neq -7-4\neq 10-(-7)$

Hence,the given sequence does not form an arithmetic progression.

$\frac{4}{-1}=-4\\\frac{-7}{4}=\frac{-7}{4}\\\frac{10}{-7}=\frac{-10}{7}\\So,\,\,\frac{4}{-1}\neq \frac{-7}{4}\neq \frac{10}{-7}$

Hence,the given sequence does not form a geometric progression.

So, $-1,4,-7,10,...$ is neither an arithmetic progression nor a geometric progression.

For $192,24,3,\frac{3}{8},...$ :

$\frac{24}{192}=\frac{1}{8}\\\frac{3}{24}=\frac{1}{8}\\\frac{\frac{3}{8}}{3}=\frac{1}{8}\\So,\,\,\frac{24}{192}=\frac{3}{24}=\frac{\frac{3}{8}}{3}$

As ratio of the consecutive terms is same, the sequence forms a geometric progression.

For $-25,-18,-11,-4,...$ :

$-18-(-25)=-18+25=7\\-11-(-18)=-11+18=7\\-4-(-11)=-4+11=7\\So,\,\,-18-(-25)=-11-(-18)=-4-(-11)$

As the difference between the consecutive terms is the same, the sequence forms an arithmetic progression.

3 0

2 years ago

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

2 years ago

Estimate the sum 8/10 + 1/9

RideAnS [48]

Answer:

41/45

Step by Step Explanation:

Add: 8/

+ 1/

= 8 · 9/

10 · 9

+ 1 · 10/

9 · 10

= 72/

+ 10/

= 72 + 10/

= 82/

= 2 · 41/

2 · 45

= 41/

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(10, 9) = 90. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 10 × 9 = 90. In the following intermediate step, cancel by a common factor of 2 gives 41/

In other words - eight tenths plus one ninth = forty-one forty-fifths.

8 0

2 years ago