Twenty percent of what number is equal to five percent of 680

Which equation can you use to solve in the figure shown?

Neporo4naja [7]

What’s the figure I couldn’t see anything here.

5 0

2 years ago

The answer for Mixed Degree Systems

Setler79 [48]

Answer:

We have to find the maximum number of solutions of each of the following system:

1)

Two distinct concentric circles:

Since, distinct concentric circles means that the two circles have same center but different radius.

That means they will never intersect each other at any point.

Ans hence we will get zero solutions.

2)

Two distinct parabolas:

Two parabolas can maximum intersect at 2 points this could be seen by the diagrams.

3)

A line and a circle.

A line and a circle can maximum have 2 solutions.

4)

A parabola and a circle.

It can have maximum two solutions it can be seen from the diagram.

4 0

2 years ago

Read 2 more answers

Jason Reads 126 pages in 6 hours. Joe reads 115 pages in 5 hours how many more pages per hour does Joe read?

Marizza181 [45]

Jason --> 126/6 = 21 pages per hour

joe --> 115/5 = 23 pages per hour

joe reads 2 more pages per hour

hope this helps :)

4 0

1 year ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$