All categories

2 years ago

Solve the following inequality algebraically. 5x + 3) +8 > 43

Mathematics

2 answers:

kompoz [17]2 years ago

4 0

⚠️Question is incomplete or incorrect then don't blame me for wrong answer ⚠️

$\huge \boxed{ \tt 6.4}$

Steps :-

(5x + 3) +8 > 43
5x + 3 + 8 >43
5x + 11 >43
5x >43 - 11
5x >32
x > 32 ÷ 5
x>32/5
x > 6.4

allochka39001 [22]2 years ago

4 0

Solution:

(5x + 3) + 8 > 43

Open the bracket.

5x + 3 + 8 > 43

Simplify the LHS.

5x + 11 > 43

Subtract 11 both sides.

5x + 11 - 11 > 43 - 11
=> 5x > 32

Divide 5 both sides.

=> x > 32/5

The simplified inequality is x > 32/5.

You might be interested in

1/2y=13 Solve for y.

8090 [49]

Answer:

The correct answer is Y = 26

7 0

3 years ago

Read 2 more answers

The product of two ratio rational numbers is 48/5. If one of the rational number is 66/7, find the other rational number.

goldenfox [79]

Let x = the other rational number.

x(66/7) = (48/5)

Solve for x to find your answer.

3 0

3 years ago

Please please please

lapo4ka [179]

The answer is y= 3/4x - 4.

6 0

4 years ago

Line segment YV of rectangle YVWX measures 24 units.

Anna [14]

Answer: 8√3

Step-by-step explanation:

ΔXVW is a 30°-60°-90° triangle. This special triangle has corresponding sides of lengths: b - b√3 - 2b. So,

VW = b

XW = b√3

XV = 2b

Since, YV = 24, then XW = 24 and we stated above that XW = b√3. So,

24 = b√3

$\frac{24}{\sqrt{3}} = \frac{b\sqrt{3}}{\sqrt{3}}$

$\frac{24}{\sqrt{3}}$ = b

$\frac{24}{\sqrt{3}}(\frac{\sqrt{3}}{\sqrt{3}})$ = b

$\frac{24\sqrt{3}}{3}$ = b

8√3 = b

We are looking for side YX which is congruent (equal) to VW and we stated above that VW = b, So, VW = 8√3

7 0

3 years ago

Use a matrix to solve the system:

Romashka-Z-Leto [24]

Answer:

(2.83 , 1 , 4)

Step-by-step explanation:

$2x+2y-z=4\\4x-2y-2z=2\\3x+3y-4z=-4\\$

Rewrite these equations in matrix form

$\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\$

we can write it like this,

$AX=B\\X=A^{-1}B$

so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.

We get the inverse of matrix A,

$A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right] \\$

now multiply the matrix with B

$X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\$

4 0

3 years ago