Sylvie finds the solution to the system of equations by graphing.

What is the mode of this data set? 4, 69, 84, 69, 90, 75, 94, 90, 90, 1, 5

nika2105 [10]

Answer:

90

Step-by-step explanation:

90 because it occurs most in the data set.

3 0

3 years ago

Read 2 more answers

Dennis and Ivy share some sweets between them in the ratio 7:4 Dennis got 42 sweets. How many does ivy get

MrRa [10]

Answer:

$Ivy = 24$

Step-by-step explanation:

Given

$Dennis : Ivy = 7 : 4$

Required

Determine the amount of sweet Ivy gets when Dennis = 42

We have:

$Dennis : Ivy = 7 : 4$ and

$Dennis = 42$

Substitute 42 for Dennis in $Dennis : Ivy = 7 : 4$

$42 : Ivy = 7 : 4$

Convert to fraction

$\frac{42}{Ivy} = \frac{7}{4}$

Cross Multiply:

$Ivy * 7 = 42 * 4$

Divide through by 7

$\frac{Ivy * 7}{7} = \frac{42 * 4}{7}$

$Ivy = \frac{42 * 4}{7}$

$Ivy = 6* 4$

$Ivy = 24$

6 0

3 years ago

So sorry to ask another question but I need some help-

Vika [28.1K]

Answer:

14x + 23

Step-by-step explanation:

The lengths of three sides of a triangle are 5x + 9 feet, 2x + 14 feet, and 7x feet.

The perimeter of a triangle: the sum of all of its sides.

5x + 9 + 2x + 14 + 7x

14x + 23

It's the expression of the perimeter of the triangle. Once you will be asked to find x, you'll probably be given the exact perimeter of the triangle.

7 0

2 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

What are the graphs of y=sinx and y=cosx in the interval from -2 pi to 2 pi?

g100num [7]

I used Excel to create the tables and the graphs.

I attached both the tables and the graphs.

You should replace the numbers of the x-axis (in both graphs) by the numbers as a fraction of pi. Those numbers are also included in the table, so you should not have problems with that.

Open and see tha file attached with the answer to your question..

Download pdf

5 0

3 years ago