Determine the AP whose 3rd term is 5 and 7th term is 9

7t-3t+4t-5t show working

sukhopar [10]

Answer:

3t

Step-by-step explanation:

7t-3t+4t-5t

Combine like terms

7t -3t = 4t and substitute back into the expression

4t +4t -5t

4t+4t = 8t and substitute back into the expression

8t - 5t

3t

7 0

2 years ago

If y varies directly as x, and y is 400 when x is randy is r when x is 4, what is the numeric constant of variation in this rela

juin [17]

The numeric constant of variation is 100

Step-by-step explanation:

Direct proportion means that increase in one variable cause increase in other variable

Here,

It is given that

y varies directly as x

y∝x

when the proportionality symbol is removed a proportionality constant is introduced

$y = kx$

Given that

y = 100 when x = 4

Putting the values

$400 = k * 4\\k = \frac{400}{4}\\k = 100$

The numeric constant of variation is 100

Keywords: Proportion variation

Learn more about direct proportion at:

#LearnwithBrainly

5 0

3 years ago

Which property of equality would be used to solve 3x = 81?

slega [8]

Division x=81/3 that’s right

4 0

3 years ago

Read 2 more answers

Need helpp fastt plss will give brainliest to who is correct plss helpp

Mandarinka [93]

1-7
3-21
5-35
7-49

Hope this helps!

6 0

2 years ago

Read 2 more answers

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:

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