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nlexa [21]
2 years ago
14

Wesley willed his estate to Margaret,

Mathematics
2 answers:
Jlenok [28]2 years ago
8 0

Answer:

1,100,000

Step-by-step explanation:

DiKsa [7]2 years ago
4 0

Answer:

1,100,000

Step-by-step explanation:

20000 + 22(20000) + 32 (20000)= 55 (20000)= 1,100,000

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What is the image point of (-8, -9) after a translation left 3 units and down 5<br> units?
dimulka [17.4K]

Answer: (-11,-14)

Step-by-step explanation:

3 units left means you subtract 3 from the x-value and 5 units down means you subtract 5 from the y-value. So -8-3=-11 and -9-5=-14. So the answer is (-11,-14)

8 0
3 years ago
The football club sold bags of candy and cookies to raise money for the spring show. Bags of candy cost $6.00
faust18 [17]

Answer:

Candy=x=2.8 bags

Cookies=3+2.8=5.8 bags

Step-by-step explanation:

Price

Bag of candy=$6.00

Bag of cookies=$4

Total sales=$40

Quantity

Let cookies=y=3+x

Let candy=x

There are 3 more bag of cookies than candy sold

Total sales=price*quantity of x + price*quantity of y

40=$4*(3+x) + 6*x

40=12+4x+6x

40=12+10x

40-12=10x

28=10x

X=28/10

x=2.8 bags

6 0
3 years ago
Rectangle abcd is similar to rectangle pqrs given that ab =14cm bc =8cm and pq=12cm calculate the length of qr
ohaa [14]

Answer: 6.8571 (Round as needed)

Hope this is correct

Step-by-step explanation:

This is a simple ratio problem

Using the similarity statement we can say 14:12, (ab:pq)

That is our ratio.

So we do 14/12 = 8/x, we solve this using algebra to get our answer.

Hope this is correct.

7 0
2 years 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.

<u>Ben:</u>

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

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

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

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>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 </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

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

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

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

3 0
3 years ago
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