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3 years ago

Lisa is 100 yards away from tim and lisa can run to tim in 18 sec

Mathematics

1 answer:

ladessa [460]3 years ago

6 0

Lisa runs approximately 5 and 5 / 9 yards per second.

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Which equations represent the line that is parallel to 3x − 4y = 7 and passes through the point (−4, −2)? Select two options.

Elanso [62]

Answer:

2 and 5

Step-by-step explanation:

The slope-intercept form of a line is y=mx-b where m is slope and b is y-intercept.

The point-slope form of a line is y-y1=m(x-x1) where m is the slope and (x1,y1) is a point on the line.

The standard form a line is ax+by=c.

So anyways parallel lines have the same slope.

So if we are looking for a line parallel to 3x-4y=7 then we need to know the slope of this line so we can find the slope of our parallel line.

3x-4y=7

Goal: Put into slope-intercept form

3x-4y=7

Subtract 3x on both sides:

-4y=-3x+7

Divide both sides by -4:

$y=\frac{-3}{-4}x+\frac{7}{-4}$

Simplify:

$y=\frac{3}{4}x+\frac{-7}{4}$

So the slope of this line is 3/4. So our line that is parallel to this one will have this same slope.

So we know our line should be in the form of $y=\frac{3}{4}x+b$ .

To find b we will use the point that is suppose to be on our new line here which is (x,y)=(-4,-2).

So plugging this in to solve for b now:

$-2=\frac{3}{4}(-4)+b$

$-2=-3+b$

$3-2=b$

$b=1$

so the equation of our line in slope-intercept form is $y=\frac{3}{4}x+1$

So that isn't option 1 because the slope is different. That was the only option that was in slope-intercept form.

The standard form of a line is ax+by=c and we have 2 options that look like that.

So let's rearrange the line that we just found into that form.

$y=\frac{3}{4}x+1$

Clear the fractions because we only want integer coefficients by multiplying both sides by 4.

This gives us:

$4y=3x+4$

Subtract 3x on both sides:

$-3x+4y=4$

I don't see this option either.

Multiply both sides by -1:

$3x-4y=-4$

I do see this as a option. So far the only option that works is 2.

Let's look at point slope form now.

We had the point that our line went through was (x1,y1)=(-4,-2) and the slope,m, was 3/4 (we found this earlier).

y-y1=m(x-x1)

Plug in like so:

y-(-2)=3/4(x-(-4))

y+2=3/4 (x+4)

So option 5 looks good too.

7 0

3 years ago

Read 2 more answers

Which of the following would be an acceptable first step in simplifying the expression sinx/1-sinx

Paha777 [63]

$\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------$

$\bf \cfrac{sin(x)}{1-sin(x)}\implies \cfrac{sin(x)}{1-sin(x)}\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \stackrel{first~step}{\cfrac{sin(x)[1+sin(x)]}{[1-sin(x)][1+sin(x)]}}
\\\\\\
\cfrac{sin(x)[1+sin(x)]}{1^2-sin^2(x)}\implies \cfrac{sin(x)[1+sin(x)]}{cos^2(x)}
\\\\\\
\cfrac{sin(x)+sin^2(x)}{cos^2(x)}\implies \cfrac{sin(x)}{cos^2(x)}+ \cfrac{sin^2(x)}{cos^2(x)}
\\\\\\
\cfrac{sin(x)}{cos(x)}\cdot \cfrac{1}{cos(x)}+\cfrac{sin^2(x)}{cos^2(x)}\implies tan(x)sec(x)+tan^2(x)
\\\\\\
tan(x)[sec(x)+tan(x)]$

8 0

3 years ago

Read 2 more answers

What is a equivalent expression for -3(7+5g)

bogdanovich [222]

Answer:

= -21 - 15g

Step-by-step explanation:

Given that:

= -3(7+5g)

By simplifying

As "-" sign will invert the inner signs:

= -21 - 15g

I hope it will help you!!

6 0

3 years ago

Alex purchased a new car for $28000. The car depreciates 7.25% each year. what will the price be in 5 years?

dalvyx [7]

Use the depreciation formula.

$\sf p(1-r)^n$

Where 'p' is the principal value, 'r' is the rate it depreciates, and 'n' is the time. Just plug in what we know:

$\sf 28000(1-0.0725)^5$

Simplify by subtracting:

$\sf 28000(0.9275)^5$

Simplify the exponent:

$\sf 28000(0.6864)$

Multiply:

$\sf\approx\boxed{\sf\$ 19219.20}$

8 0

3 years ago

Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th

zalisa [80]

We have the following equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8$

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

$y \leq -0.5x+5 \\ y \leq -1.25x+8$

Region B.

Let's take a point that is in this region, that is:

$P(0,6)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \leq -1.25x+8$

Region C.

A point in this region is:

$P(0,10)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \ \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Region D.

A point in this region is:

$P(8,0)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \ \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2$

So the inequalities are:

$(1) \ y \leq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\ \\ Solving \ for \ y: \\ y=-0.5(4)+5=3$

Accordingly, the point is:

$\boxed{p(4,3)}$

Point q:

This point is the $x-intercept$ of the line:

$y=-0.5x+5$

So let:

$y=0$

Then

$x=\frac{5}{0.5}=10$

Therefore, the point is:

$\boxed{q(10,0)}$

Part b)

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

$O(0,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}$

It is true.

Region B.

Let's take a point that is in this region, that is:

$P(0,7)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}$

It is true

Region C.

Let's take a point that is in this region, that is:

$P(0,11)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}$

It is true

Region D.

Let's take a point that is in this region, that is:

$P(9,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}$

It is true

7 0

4 years ago