Please help asap! i will give brainliest.

❗️‼️❗️HELP❗️‼️❗️ Harry had a pile of 48 pennies. He organized them into a rectangular array with exactly four rows with 12 penni

Viefleur [7K]

Answer:

One array you could make is 2 rows of pennies with 24 pennies in each row. Another array you could make is 3 rows of pennies with 16 pennies in each row.

Yet another array you could make is 6 rows of pennies with 8 pennies in each row.

You could also switch all of these arrays around. For example:

- 24 rows with 2 pennies in each row

- 12 rows with 4 pennies in each row

- 8 rows with 6 pennies in each row

- 16 rows with 3 pennies in each row

8 0

3 years ago

Can anyone please help me on this thank you

Kay [80]

So are the bottom numbers the answers? Becasue if the top are the questions then gallon would be largest, 2pt, then 0.95 liter If this doesn't help let me know because the problem doesn't look complete. Let me know!

5 0

4 years ago

A bank offers all savings accounts 5% interest compounded annually. If one account has a principal of $100 and another

arsen [322]

Answer:

From the calculation for both the accounts, it is clear that both the account double in the same time period of 14 years 26 days .

Step-by-step explanation:

Given as :

The principal for the first account = p = $100

The rate of interest = r = 5% compounded annually

The account gets double , so, Amount = A = $200

Let the time after which account gets double = t years

So, From Compound Interest method

Amount = Principal × $(1+\dfrac{\trxtrm rate}{100})^{\textrm time}$

As amount is double its principal

So, A = 2 × $100 = $200

Or, A = p × $(1+\dfrac{\trxtrm r}{100})^{\textrm t}$

Or, $200 = $100 × $(1+\dfrac{\trxtrm 5}{100})^{\textrm t}$

Or, $\dfrac{200}{100}$ = $(1.05)^{\textrm t}$

Or, 2 = $(1.05)^{\textrm t}$

Taking Log both side

$Log_{10}2$ = $Log_{10}(1.05)^{t}$

Or, 0.3010 = t $Log_{10}1.05$

Or, 0.3010 = t × 0.0211

∴ t = $\dfrac{0.3010}{0.0211}$

I.e t = 14.26

So, The time period to get account double is 14 years 26 days

Again

Amount = Principal × $(1+\dfrac{\trxtrm rate}{100})^{\textrm time}$

Or, A = p × $(1+\dfrac{\trxtrm r}{100})^{\textrm t}$

As amount is double its principal

So, A = 2 × $1000 = $2000

Or, $2000 = $1000 × $(1+\dfrac{\trxtrm 5}{100})^{\textrm t}$

Or, $\dfrac{2000}{1000}$ = $(1.05)^{\textrm t}$

Or, 2 = $(1.05)^{\textrm t}$

Taking Log both side

$Log_{10}2$ = $Log_{10}(1.05)^{t}$

Or, 0.3010 = t $Log_{10}1.05$

Or, 0.3010 = t × 0.0211

∴ t = $\dfrac{0.3010}{0.0211}$

I.e t = 14.26

So, The time period to get account double is 14 years 26 days

Hence From the calculation for both the accounts, it is clear that both the account double in the same time period of 14 years 26 days . Answer

7 0

3 years ago

Can someone help me with this question, please. The problem is attached.

fomenos

It is a reflection across the y axis, then a translation to the right 2 units.

I hope this helps! If you have more you want help with I can help.

3 0

3 years ago

Which graph correctly represents 1/3y-1/2x>2 ?

Tanzania [10]

Put it in the form of y =mx +b, or in this instance, y> mx +b
move the (1/2) x to the right by adding it to both sides of the inequality
(1/3)y>(1/2)x +2
Multiply by 3 on both sides to get y by itself.
y>(3/2) x +6
This is a graph with y intercept of (0,6) and a moderate upward and to the right slope. Because it is > , the line on the graph will NOT be part of the solution.

The easiest way to find the side of the graph that the inequality satisfies is to use (0,0) and see if it works or doesn't work. In the original equation, 0-0>2 does NOT work, so the area where the inequality works is to the up and left of the graph, which should be a dotted line to show that the inequality is greater than only.
The point (6,-2) should work.
Test it. 6*(1/3)-(-2)*(1/2)>0 ; 2-(-1)=3, and 3>2 It does work.
The point (6,2) should not work
Test it. 6 *(1/3)-2(1/2)=2-1 ; 1 is NOT >2, so it does not work.
If the graph goes through the origin, then pick a point near the graph with a small x or y.

7 0

4 years ago