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3 years ago

Relationship B has a greater rate than Relationship A. The graph represents Relationship A.

Mathematics

2 answers:

Musya8 [376]3 years ago

8 0

What was the answer????

Pepsi [2]3 years ago

7 0

I don’t understand. Could you break the question down?

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A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns shown, making the

Leno4ka [110]

Answer:
The original selling price is $520.632
Explanation:
Consider the initial selling price be $x
After marking down 10%,
Final Selling price = Selling price(1-10%)
Selling price = x – 10% of x = x – 0.1x = 0.9x
After marking 30%
Final Selling price = Selling price(1-10%)
Selling price = 0.9x – 0.3*0.9x = 0.9x – 0.27x = 0.63x
According to the question,
0.63x = 328

Therefore, the original selling price is $520.632

5 0

3 years ago

Read 2 more answers

write the equation of a line in slope intercept form that contains the points ( - 5,2 ) and ( 1, - 4 )

Natalija [7]

Equation of a line is given by y - y1 = m(x - x1); where m = (y2 - y1)/(x2 - x1)

y - 2 = ((-4 - 2)/(1 - (-5))(x - (-5))
y - 2 = (-6/(1 + 5))(x + 5)
y - 2 = (-6/6)(x + 5)
y - 2 = -1(x + 5)
y - 2 = -x - 5
y = -x - 5 + 2
y = -x - 3

8 0

3 years ago

The functions f(x)=−34x+214 and g(x)=(12)x+1 are shown in the graph. What numbers are solutions to −34x+214=(12)x+1? Select each

DENIUS [597]

Answer:

$x=-1$

$x=1$

Step-by-step explanation:

we have

$f(x)=-\frac{3}{4}x+2\frac{1}{4}$ -----> equation A

$g(x)=\frac{1}{2}^{x} +1$ ----> equation B

To find the solutions for x of the system of equations

equate equation A and equation B

$-\frac{3}{4}x+2\frac{1}{4}=\frac{1}{2}^{x} +1$

The values of x of the intersection points are the solutions of the system

Using a graphing tool

see the attached figure

There are two intersection points

therefore

The solutions are

$x=-1$

$x=1$

8 0

3 years ago

In April, the number of cars sold was 546.

Maksim231197 [3]

Answer:

520 cars

Step-by-step explanation:

Let the car sold in March be x, x+5%*x=546. (105/100)*x=546. x=520

4 0

3 years ago

The height, h, in meters above the ground, of a projectile at any time, t, in seconds, after the launch is defined by the functi

Ivanshal [37]

The equation $h(t) =-6t^2 + 15t + 2$ sets the parabola. You can find the parabola vertex as:

$t_v= \frac{-b}{2a} =- \frac{15}{2\cdot (-6)} = \frac{15}{12}= \frac{5}{4}=1.25$

then

$h( \frac{5}{4} )=-6\cdot (\frac{5}{4} )^2+15\cdot \frac{5}{4} +2= \frac{91}{8} =11.375$

This means that the maximum height is 11.375 m and a projectilereaches this height at 1.25 seconds.

The projectile initially launched at time t=0, so the initial height is h(0)=2 m

6 0

4 years ago