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3 years ago

A) equal lines B) parallel lines C) perpendicular lines D) None of the above

Mathematics

2 answers:

Nat2105 [25]3 years ago

8 0

I think the answer is D, None of the above

navik [9.2K]3 years ago

8 0

Answer:

D its none of the above answers

Step-by-step explanation:

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How do I complete this formal check?

natima [27]

Answer:

Step-by-step explanation:

The first two steps are for the purpose of eliminating fractions. Doing so results in 4(2x - 5) = 9(x - 2), which is to be solved for x.

Perform the indicated multiplication, obtaining:

8x - 20 = 9x - 18.

Then combine like terms: -2 = x, or

x = -2.

To complete the formal check, substitute -2 for x in the first equation. This results in:

(1/3)(-4 - 5) = (3/4)(-2 -2), or

-3 = -3

... which is obviously true.

8 0

3 years ago

Jefferson Middle School needs to see a 300% gain in PTO fundraiser

Ronch [10]

Given parameters:

Percentage gain = 300%

Selling cost last year = $2420

Unknown:

Selling cost this year = ?

Solution:

To solve this problem;

Profit percent = $\frac{Selling cost this year - Selling cost last year}{Selling cost last year} x 100$

300 = $\frac{S - 2420}{2420} x 100$

3 = $\frac{S - 2420}{2420}$

S- 2420 = 7260

S = $9680

To make a 300% gain, Jefferson Middle school needs a $9680 from fundraiser sales this year.

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3 years ago

What is the approximate circumference of the circle shown below?

Serggg [28]

Hey, there’s no circle for me to look at unfortunately!

4 0

3 years ago

What is (2 × 2) to the 2nd power, minus 4?

Ivan

Well first, you would do 2 x 2 which equals 4. Then you make 4 to the 2nd power, which is 16. You subtract 4 from 16, and it's 12.

7 0

3 years ago

Read 2 more answers

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

3 years ago