Select the inequality below that could have 7 as a solution.

Multiply: <br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B8n%5E%7B2%7D%20%7D%7Bm%5E%7B2%7D-16%20%7D" id="TexFormula1" title="\f

Mkey [24]

9514 1404 393

Answer:

4mn/(3m+12)

Step-by-step explanation:

It is often helpful to factor expressions so that common factors can cancel.

$\dfrac{8n^2}{m^2-16}\times\dfrac{m^2-4m}{6n}=\dfrac{8mn^2(m-4)}{6n(m-4)(m+4)}=\dfrac{4mn}{3(m+4)}\\\\=\boxed{\dfrac{4mn}{3m+12}}$

3 0

2 years ago

What will happen if pressure is increased?

Anika [276]

Answer:

the equilibrium will shift towards the side

3 0

3 years ago

1. Michael buys a chocolate bar which has 50% extra free. The chocolate bar usually weighs 200g. (a) How much extra chocolate do

KiRa [710]

Answer:

1/2 more chocalte or 50% more or 100g more

300g is the new weight

Step-by-step explanation:

200/2=100(50%)

100+200=300g

5 0

2 years ago

The given line segment has a midpoint at (3, 1).

Katena32 [7]

Answer:

$y=\frac{1}{3}x$

Step-by-step explanation:

The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)

Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through $(x_1,y_1)\ and\ (x_2,y_2)$ is:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Therefore the line passing through (2, 4) and (4, -2) has an equation:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10$

Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10

Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:

$m_1m_2=-1$ .

Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:

$m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}$

Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:

$\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x$

the equation, in slope-intercept form, of the perpendicular bisector of the given line segment is $y=\frac{1}{3}x$

7 0

3 years ago

Read 2 more answers

If TV ≅ WV and UV ≅ XV, then which congruency theorem shows that ΔTUV ≅ ΔWXV ?

VashaNatasha [74]

Answer:

Option (2).

Step-by-step explanation:

In the figure attached,

Given : TV ≅ WV and UV ≅ XV

To prove: ΔTUV ≅ ΔWXV

Statements Reasons

1). TV ≅ WV 1). Given

2). UV ≅ XV 2). Given

3). ∠XVW ≅ ∠TVU 3). Reflexive property

4). ΔTUV ≅ ΔWXV 4). SAS (Side - angle - side) property of congruence

Therefore, Option (2) will be the answer.

5 0

3 years ago