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3 years ago

Arrange the pennies into a complete rectangular array and then into a different rectangular array that has a reminder of 1

Mathematics

1 answer:

Inessa [10]3 years ago

4 0

I'm guessing the answer would be 141 pennies in a rectangular array I think I'm right but i might be wrong

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When solving problems with units we sometimes ____ to get the units the same.

laiz [17]

Convert is your answer .-.

4 0

3 years ago

Read 2 more answers

Drag each tile to the correct box.

Natasha_Volkova [10]

Answer:

1) Function h

interval [3, 5]

rate of change 6

2) Function f

interval [3, 6]

rate of change 8.33

3) Function g

interval [2, 3]

rate of change 9.6

Step-by-step explanation:

we know that

To find the average rate of change, we divide the change in the output value by the change in the input value

the average rate of change is equal to

$\frac{f(b)-f(a)}{b-a}$

step 1

Find the average rate of change of function h(x) over interval [3,5]

Looking at the third picture (table)

$f(a)=h(3)=4$

$f(b)=h(5)=16$

$a=3$

$b=5$

Substitute

$\frac{16-4}{5-3}=6$

step 2

Find the average rate of change of function f(x) over interval [3,6]

Looking at the graph

$f(a)=f(3)=10$

$f(b)=f(6)=35$

$a=3$

$b=6$

Substitute

$\frac{35-10}{6-3}=8.33$

step 3

Find the average rate of change of function g(x) over interval [2,3]

we have

$g(x)=\frac{1}{5}(4)^x$

$f(a)=g(2)=\frac{1}{5}(4)^2=\frac{16}{5}$

$f(b)=g(3)=\frac{1}{5}(4)^3=\frac{64}{5}$

$a=2$

$b=3$

Substitute

$\frac{\frac{64}{5}-\frac{16}{5}}{3-2}=9.6$

therefore

In order from least to greatest according to their average rates of change over those intervals

1) Function h

interval [3, 5]

rate of change 6

2) Function f

interval [3, 6]

rate of change 8.33

3) Function g

interval [2, 3]

rate of change 9.6

7 0

3 years ago

X²+y²-6x+14y-1=0<br> and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

$x =\frac{-b+\sqrt{b^2}-4ac }{2a}$ , $\frac{-b-\sqrt{b^2} -4ac }{2a}$

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

3 0

2 years ago

an article after allowing a discount of 20% on its marked price was sold at a gain of 20% had it been sold after allowing 25% di

Effectus [21]

Answer:

Rs 120

Step-by-step explanation:

Let MP = m

Discount = 25% of MP = m/4

SP = MP - Discount

= m - m/4 = 3m/4

If profit is 20%, CP = 3m/4 x 100/120 = 5m/8

=> Profit = 3m/4 - 5m/8 = m/8

Given Profit - Discount = Rs 15

=> |m/8 - m/4| = 15

m/8 = 15

m = Rs 120

8 0

3 years ago

a school paid $31.50 for each calculator, if the school spent 504 dollars on calculators how many did the school buy?

monitta

The school bought 16 calculators.

504/31.50 =16

7 0

3 years ago