PLEASE HELP!!<br> Solve each inequality and graph its solution.

I NEED HELP WITH MY HOMEWORK SOMEONE HELP PLS

mr Goodwill [35]

Answer:

No

Step-by-step explanation:

The question is asking for difference which is another term for subtraction.In that regard Amy is correct, however the question asks for the difference between 3 and -5 which means she'd have to subtract 3 from -5 not the other way around.

4 0

2 years ago

1. Giles thinks that -5x²(3x - 4) simplifies to 15x² + 20x². Explain what mistakes he made. What is the

jeka57 [31]

Answer:

The mistake was minus sign before the 15 and putting 15x² instead of 15x³.

The correct answer is:

$-5x^{2} (3x-4)\\ = -15x^{3} + 20x^{2}$

6 0

3 years ago

£310 is divided between Angad, Nick & June so that Angad gets twice as much as Nick, and Nick gets three times as much as Ju

Sonja [21]

nick gets 1800 pounds. Hope it helps

4 0

3 years ago

Read 2 more answers

The line segment joining the points ( 3,-4) and (1,2) is trisected at the points P and Q . If the coordinates of P and Q are (p

sdas [7]

Answer:

P = 7/3 and q = 0

Step-by-step explanation:

The given parameters are;

The endpoints of the segment are, (3, -4) and (1, 2)

The points that trisect the given points are P and Q with coordinates (p, -2) and (5/3, q) respectively

Therefore, we have;

The first point of the trisection cuts 1/3 of the length from one point and the second point of the trisection cuts 2/3 of the length from the same point

The coordinates of P or Q = (3 - (3 - 1)/3, -4 - (-4 - 2)/3) = (7/3 , -2)

Therefore, given that the y-coordinate value of the derived point coincides with the y-coordinate value of the point P, (p, -2) and there is only one point with x = -2 on the line, we have that the coordinate of the point P is (p, -2) = (7/3 , -2)

∴ P = 7/3

Similarly we have the second point of the trisection, Q, given as follows;

We are already given the x-coordinate value of the point Q as the 5/3 in (5/3, q)

Point Q = (5/3, q) = (3 - 2×(3 - 1)/3, -4 - 2×(-4 - 2)/3) = (5/3 , 0)

Point Q = (5/3, q) = (5/3 , 0)

∴ q = 0.

3 0

3 years ago

Write each expression as an algebraic (nontrigonometric) expression in u, u > 0.

max2010maxim [7]

Answer:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0$

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

Take the secant of both sides:

$\displaystyle \sec(\theta)=\frac{u}{10}$

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

$\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}$

By substitutition:

$\displaystyle= \sin(2\theta)$

Using an double-angle identity:

$=2\sin(\theta)\cos(\theta)$

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

4 0

3 years ago

PLEASE HELP!! Solve each inequality and graph its solution.