How do you find the question mark on #4?

How to write 28 divided by 7 as an expression

Korolek [52]

Answer:

28/7

Step-by-step explanation:

5 0

3 years ago

Tonya and Leo each bought a cell phone at the same time. The trade-in values, in dollars, of the cell phones are modeled by the

Paha777 [63]

Answer:

The answer is Tonya's phone had the greater initial trade-in value.

Leo's phone decreases at an average rate slower than the trade in value of Tonya's phone.

Step-by-step explanation:

Given

Tonya

$f(x) = 490 * 0.88^x$

Leo

$x \to g(x)$

$0 \to 480$

$2 \to 360$

$4 \to 470$

Solving (a): The phone with greater initial value

The initial value is when x = 0. So, we have:

$f(x) = 490 * 0.88^x$

$f(0) = 490 * 0.88^0$

$f(0) = 490 * 1$

$f(0) = 490$

From Leo's table

$g(0) = 480$

By comparison;

$f(0) > g(0)$

i.e.

$490 > 480$

<em>So: Tonya's had the greater initial trade-in value</em>

Solving (b): The phone with lesser rate

An exponential function is:

$y = ab^x$

Where:

$b \to rate$

For Tonya

$b = 0.88$

For Leo, we have:

$(x_1,y_1) = (0,480)$

$(x_2,y_2) = (2,360)$

So, the equation becomes:

$y = ab^x$

$480 = ab^0$ and $360 = ab^2$

Solving $480 = ab^0$ , we have:

$480 = a * 1$

$480 = a$

$a= 480$

$360 = ab^2$ becomes

$360 = 480 * b^2$

Divide both sides by 480

$0.75 = b^2$

Take square roots

$0.87 = b$

$b=0.87$ -- Leo's rate

By comparison; Leo's rate is slower i.e. 0.87 < 0.88

6 0

3 years ago

Brian bought a 15 pound bag of comcrete. he split the bag into smaller bags that each hold two and a half pounds. how many small

Westkost [7]

Okay. You have 15 lbs of concrete, and x bags that can individually hold 2.5 lbs. What you need to do is divide 15 by 2.5, and the quotient will be 6.
Problem: 15 ÷ 2.5 = 6
So the answer to your problem will be: 6 bags.

3 0

3 years ago

Read 2 more answers

Two dogs run around a circular track 300 m long. One dog runs at a steady rate of 15m per second, the other at a steady rate of

mihalych1998 [28]

Answer:

300 seconds

Step-by-step explanation:

The first dog run at v₁ = 15 m/sec the second one run at v₂ = 12 m/sec

we know that d = v*t then t = d/v

Then the first dog will take 300/ 12 = 25 seconds to make a turn

The second will take 300 / 15 = 20 seconds to make a turn

Then the first dog in 12 turns 12*25 will be at the start point, and so will the second one at the turn 15.

To check first dog 12 * 25 = 300

And the second dog 15 * 20 = 300

That means that time required for the two dogs to be at the start point together is

300 seconds, in that time the first dog finished the 12 turns, and the second had ended the 15.

Another procedure to solve this problem is as follows:

between 12 m/sec and 15 m/sec the minimum common multiple is 300 ( 300 is the smaller number that accept 12 and 15 as factors 12*15 = 300) Then when time arrives at 300 seconds the two dogs will be again in the starting point

3 0

3 years ago

I need help with all of them

sveticcg [70]

I just answered this xD

1-3 is given as an example.

4. True

5. False. It is the same thing twice

6. True

3 0

3 years ago