All categories

3 years ago

Shari had six dimes and nine nickels.Did she have enough money to buy a $0.99 pack of gum

Mathematics

2 answers:

Flura [38]3 years ago

8 0

Answer:yes she does cauce she has 1.05

Step-by-step explanation:

pochemuha3 years ago

7 0

Answer:

yes

Step-by-step explanation:

she has $1.05

You might be interested in

Simplify 3^2 x 3^3 <br><br> A. 3<br><br> B. 3^6<br><br> C. 3^5<br><br> D. 3^8

Maslowich

When you are multiplying and have like bases, then you add the exponents.
2+3=5
3^5

5 0

2 years ago

HELP LAST PROBLEM!

Keith_Richards [23]

Given 4 boxes of pencils cost $5,
Unknown- slope
$1.25 per box since $5/boxes
Equation y=mx+b
Substitute $1.25/per box for m
Solve y=$1.25x
The Check
$5=$1.25/per box(4 boxes)
$5=$5

3 0

2 years ago

Someone help please

Rzqust [24]

Answer:

A=(b^2)h/2

Step-by-step explanation:

7 0

2 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

3 years ago

Helppppp ASAP thankkssss

Tcecarenko [31]

The answer is 8 DVDs per stack

7 0

2 years ago