Help me ASAP<br> Simplify. <br> 7 – 2 • 5 + 9 • 3<br> A.24<br> B.102<br> C.-30<br> B.18

A baker buys a 5lb bag of suger. She uses 1 and 2/3 lb to make muffins and 2 and 3/4 lb to make a cake. How much suger does she

Aleonysh [2.5K]

Answer: 7/12 lbs

A baker buys a 5 lb bag of sugar. She uses 1 2/3 lb to make some muffins and 2 3/4 lb to make a cake. How much sugar does she have left?

6 0

3 years ago

Read 2 more answers

How many proportional relationships are shown in the coordinate plane below?

aev [14]

Answer:

C: 2

Step-by-step explanation:

${ \purple{purple \: line}} \: \: and \: \: { \green{green \: line}}$

3 0

3 years ago

Find a sum equivalent to the product shown.<br> 4(z + x)

Sergeeva-Olga [200]

Answer:

4z+4x

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

There are 112 seats in the school auditorium. There are 7 seats in each row. There are 70 people seated, filming up full rows of

AleksandrR [38]

6 rows are empty.

112/7 = 16 rows

70/7 = 10 rows used.

4 0

3 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

Help me ASAP Simplify. 7 – 2 • 5 + 9 • 3 A.24 B.102 C.-30 B.18