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oksano4ka [1.4K]

3 years ago

12

Two classes at Liberty Middle School are setting up imaginary banks for an economics project. Class 1 has $50 in a savings accou

nt and makes a $5 deposit each day. Class 2 has $110 in a savings account and makes a $5 withdrawal each day. After how many days will both classes have the same amount of money in their savings accounts?

1 answer:

ivolga24 [154]3 years ago

6 0

In 2 days, class 1 has 60 and class 2 has 100, in 4 days, class 1 has 70 and class 2 will have 90, in 6 days both classes will have 80. so 6 days is the answer

You might be interested in

The table represents a linear function. find the values of a,b, and c. Show your work.

viva [34]

Answer:

The values of a , b , c are ⇒ a = 1 , b = 10 , c = 9

Step-by-step explanation:

* Lets describe the meaning of the linear function

- Linear function is represented by a line graphically

- The equation of the line is y = mx + c, where m is the slope

of the line and c is the y-intercept (the point of intersection

between the line and the y-axis is (0 , c))

- m = change of y/change of x

- We can find m from any two points on the line

* Lets use m to find a, b and c

- Use the points (3 , 8) and (5 , 9) to find m

∵ m = (y2 - y1)/(x2 - x1)

∴ m = (9 - 8)/(5 - 3) = 1/2

- Find a by using points (a , 7) and (3 , 8) ⇒ (or (5 , 9))

∵ m = (8 - 7)/(3 - a) = 1/2

∴ 1/(3 - a) = 1/2 ⇒ by using cross multiplication

∴ 3 - a = 2 ⇒ subtract 3 from both sides

∴ -a = -1 ⇒ × -1 both sides

∴ a = 1

- Find b by using points (5 , 9) and (7 , b) ⇒ (or (3 , 8))

∵ m = (b - 9)/(7 - 5) = (b - 9)/2

∴ (b - 9)/2 = 1/2 ⇒ by using cross multiplication

∴ 2(b - 9) = 2 ⇒ open the bracket

∴ 2b - 18 = 2 ⇒ add 18 to both sides

∴ 2b = 20 ⇒ ÷ 2

∴ b = 10

- Find b by using points (5 , 9) and (c , 11) ⇒ (or (3 , 8))

∵ m = (11 - 9)/(c - 5) = 2/(c - 5)

∴ 2/(c - 5) = 1/2 ⇒ by using cross multiplication

∴ c - 5 = 4 ⇒ add 5 to both sides

∴ c = 9

* The values of a , b , c are ⇒ a = 1 , b = 10 , c = 9

6 0

3 years ago

Someone please help explain how to find x

dimaraw [331]

Answer:

x=27

Step-by-step explanation:

180-2x-40+180-3x+180-x-50+180-4x = 360

360-2x-40-3x+360-x-50-4x= 360

360-10x-90= 0

10x= 270

x= 27

5 0

4 years ago

Trouble finding arclength calc 2

kiruha [24]

Answer:

$S\approx1.1953$

Step-by-step explanation:

So we have the function:

$y=3-x^2$

And we want to find the arc-length from:

$0\leq x\leq \sqrt3/2$

By differentiating and substituting into the arc-length formula, we will acquire:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx$

To evaluate, we can use trigonometric substitution. First, notice that:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx$

Let's let y=2x. So:

$y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx$

We also need to rewrite our bounds. So:

$y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0$

So, substitute. Our integral is now:

$\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Let's multiply both sides by 2. So, our length S is:

$\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

$y=a\tan(\theta)$

Substitute 1 for a. So:

$y=\tan(\theta)$

Differentiate:

$y=\sec^2(\theta)\, d\theta$

Of course, we also need to change our bounds. So:

$\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0$

Substitute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta$

The expression within the square root is equivalent to (Pythagorean Identity):

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta$

Simplify:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta$

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

$u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta$

And:

$dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)$

Integration by parts:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)$

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)$

Distribute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)$

Now, let's make the single integral into two integrals. So:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Distribute the negative:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

$\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Divide the second and third equation by 2. So: $\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))$

Therefore, our arc length will be equivalent to:

$\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Divide both sides by 2:

$\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Evaluate:

$S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))$

Evaluate:

$S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))$

Simplify:

$S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}$

Use a calculator:

$S\approx1.1953$

And we're done!

7 0

3 years ago

ABCD is a kite.

stepan [7]

Answer:

wouldn't it be 40 if the opposite side is 40 ?

3 0

4 years ago

Read 2 more answers

Can anyone help me with these questions plz

KonstantinChe [14]

Answers:

7: 20 miles

8: 28 miles

9: 1 error

10: Their speed is 50

11: 365 words in 5 mins

Step-by-step explanation:

#7 and #8

Our equation is: $f=2.25+0.20(m-1)$ With f meaning "fare" (price) and m meaning "miles".

Question 7 - Juan's fare for his ride costs $6.05. We must solve for $m$ .

Step 1. Substitute - <em>Substitute the fare for $f$ in the equation.</em>

$6.05=2.25+0.20(m-1)$

Step 2. Simplify/Solve - Solve for $m$ .

$6.05=2.25+0.20(m-1)$

<em>- Distribute</em>

$6.05=2.25+0.2m-0.2$

<em>- Subtract 0.2 from 2.25</em>

$6.05=2.05+0.2m$

<em>- Subtract -2.05 from 6.05</em>

$4=0.2m$

<em>- Divide both sides by 0.2</em>

$20=m$

<u>And you have your answer of 20 miles.</u>

Question 8 - Same equation, different fare.

Step 1. Substitute

$7.65=2.25+0.20(m-1)$

Step 2. Solve

$7.65=2.25+0.20(m-1)\\7.65=2.25+0.20m-0.20\\7.65=2.05+0.20m\\5.6=0.20m\\28=m$

And like so, we have $m = 28$

Questions 9 - 11

<em>The equation given is $S=\frac{1}{5} (w-10e)$ . S = typing speed, w = words per 5 mins, and e= errors.</em>

Question 9 -

<em>We are given this information: S = 55, W = 285. We are solving for e</em>.

Substitute -

$55=\frac{1}{5} (285 -10e)$

Solve -

$55=\frac{1}{5} (285 -10e)\\55=\frac{1}{5} (285 -10e)\\55= 57-2e\\-2=-2e\\1=e$

So they would make 1 error.

Question 10 -

<em>Information given: 300 = w, 5=e. We are solving for S</em>

Substitute -

$S= \frac{1}{5} (300-10(5))$

Solve -

$S= \frac{1}{5} (300-10(5))\\S= \frac{1}{5} (300-50)\\S= \frac{1}{5} (300-50)\\S= \frac{1}{5} (250)\\S= 50$

Their speed is 50.

Question 11 -

<em>Information given: S = 65, e = 4. We are solving for w.</em>

Substitute -

$65=\frac{1}{5} (w-10(4))$

Solve -

$65=\frac{1}{5} (w-10(4))\\65=\frac{1}{5} (w-40)\\65=\frac{1}{5}w-8\\73=\frac{1}{5}w\\365=w$

So w = 365.

Hoped this helped!~

7 0

3 years ago