All categories

3 years ago

Please solve -4(x+10) is greater than or equal to -7x + 65

Mathematics

1 answer:

Novay_Z [31]3 years ago

7 0

Answer:

x>=35

Step-by-step explanation:

-4(x+10)>= -7x + 65

-4x-40>= -7x+65

-4x+7x>=65+40

3x>=105

x>=105/3

x>=35

You might be interested in

3x-8y=24 <br> y=x-8<br> Please show <br> work

In-s [12.5K]

Answer:

3x - 8y = 24 --------(1)

y = x - 8 ---------(2)

substitute y in (1)

3x - 8( x - 8) = 24

3x - 8x + 64 = 24

-5x = 24 - 64

-5x = -40

5x = 40

x= 8

substitute x in (2)

y = 2 - 8

y = - 6

4 0

3 years ago

Read 2 more answers

Rodrigo made 6/8 of a pound of trail mix. If he puts 1/4 of a pound into each bag, how many bags can Rodrigo fill?

ValentinkaMS [17]

Answer:

He can fill 3 bags

Step-by-step explanation:

You would first make them have a common denominator so you would have 2/8 instead of 1/4 then you would do 6/8 divided by 2/8 to get 3

1/4 = 2/8

6/8 divided by 2/8 = 3

7 0

2 years ago

Read 2 more answers

Look at the figure above. The surface area is _____ square cm.

user100 [1]

Answer: 252

Step-by-step explanation:

6 0

3 years ago

Find the value of x. Please help. I am confused

Oxana [17]

Answer:

x=30

Step-by-step explanation:

Exterior Angle Theorem = the exterior angle of a triangle is equal to the sum of the two opposite interior angle

Thus, 2x+6+36=4x-18

2x+42=4x-18 add 18 to both side

2x+60=4x minus 2x

60=2x

x=30

7 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