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2 years ago

Please help I don’t understand please!!!

Mathematics

2 answers:

Gennadij [26K]2 years ago

8 0

the (y) and mean 1 and you need to multiply al number

Vaselesa [24]2 years ago

7 0

Answer:

y>5.5

Step-by-step explanation:

4y+3>2y+14

Subtract 2y from both sides.

2y+3>14

Subtract 3 from both sides.

2y>11

Divide 2 by both sides.

y>5.5

-hope it helps

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Which of the following expressions has the greatest value?

zimovet [89]

I believe the answer is B: 3 + 7 + (-10).

4 0

3 years ago

Read 2 more answers

30° 8 + 6x 4x + 2 in a triangle

andrey2020 [161]

Answer:

x=12

Step-by-step explanation:

Simplifying

30 + 4x + 2 = 8 + 6x

Reorder the terms:

30 + 2 + 4x = 8 + 6x

Combine like terms: 30 + 2 = 32

32 + 4x = 8 + 6x

Solving

32 + 4x = 8 + 6x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-6x' to each side of the equation.

32 + 4x + -6x = 8 + 6x + -6x

Combine like terms: 4x + -6x = -2x

32 + -2x = 8 + 6x + -6x

Combine like terms: 6x + -6x = 0

32 + -2x = 8 + 0

32 + -2x = 8

Add '-32' to each side of the equation.

32 + -32 + -2x = 8 + -32

Combine like terms: 32 + -32 = 0

0 + -2x = 8 + -32

-2x = 8 + -32

Combine like terms: 8 + -32 = -24

-2x = -24

Divide each side by '-2'.

x = 12

Simplifying

x = 12

4 0

2 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

NO SOLUTION

7 0

2 years ago

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Anit [1.1K]

Answer:

The answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

Step-by-step explanation:

$\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}$

Solve the L.H.S part:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]$

After calculating the L.H.S part compare the value with R.H.S:

$\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\$

$\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\$

In equation (i) multiply by 3 and subtract by equation (iii):

$\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to c= - \frac{1}{2}$

put the value of c in equation (i):

$\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\$

In equation (ii) multiply by 3 then subtract by equation (iv):

$\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\$

put the value of d in equation (iv):

$\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2$

The final answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

4 0

3 years ago

Find all values of x that make the triangles congruent. Explain your reasoning.

RSB [31]

Answer:

x = $\frac{4}{5},5$