I would love some help :)

Mathematics

1 answer:

RUDIKE [14]3 years ago

5 0

4x+6=3x+10 so x=4, to find the measure of KJL, plug in x into the equation 4x+6. angle KJL=22°

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Nick has $7.00. Bagels cost $0.75 each, and a small container of cream cheese costs $1.29. Write an inequality to find the numbe

larisa86 [58]

The answer is 7.00=.75x+1.29 X represents number of bagels Nick can buy

3 0

3 years ago

What is the solution to the inequality <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bf%7D%7B2%7D" id="TexFormula1" title="\frac{f}

tester [92]

The solution of inequality $\frac{f}{2} - 19 < 47$ is f < 132

Solution:

Given that we have to find the solution of inequality

These things do not affect the direction of the inequality:

Add (or subtract) a number from both sides.

Multiply (or divide) both sides by a positive number.

Simplify the sides

Given inequality is:

$\frac{f}{2} - 19 < 47$

Add 19 to both sides of equation

$\frac{f}{2} - 19 +19 < 47 + 19\\\\\frac{f}{2} < 47 + 19\\\\\frac{f}{2} < 66$

Multiply both the sides of inequality by 2

$\frac{f}{2} \times 2 < 66 \times 2\\\\f < 132$

Thus the solution to given inequality is f < 132

4 0

3 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E to the system of inequalities from part A, and check if its is true, what we get.

$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to our system of inequalities.

Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).