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Alik [6]
4 years ago
14

Whats the solution to 5x+4>11-2x

Mathematics
1 answer:
aleksandrvk [35]4 years ago
8 0
The solution to the problem is x<1
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Simplify negative 15.5 divided by 5.
Naya [18.7K]

Divide:

-15.5/5 = -3.1

-3.1 is your answer

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4 years ago
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A shoe box in the shape of a rectangular prism has a length of 12 inches, a width of 6 1/2 inches, and a height of 4 1/2 inches.
katovenus [111]
12 x 6 1/2 x 4 1/2 = 351in^3
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4 years ago
fresh+cranberries+have+99%+of+water.+partially+dried+cranberries+have+98%+of+water.+jack+picked+100+pounds+of+cranberries.+how+m
Mnenie [13.5K]

Answer:

  50 lb

Step-by-step explanation:

Before drying, the cranberries have ...

  (100% -99%)(100 lb) = 1 lb

of "meat."

After drying, that represents (100% -98%) = 2% of the total weight. Then the total weight is ...

  1 lb = 2%·x

  x = (1 lb)/0.02 = 50 lb

The dried cranberries will weigh 50 lb.

8 0
2 years ago
Find the equations of the tangent lines to x2+y2=1 at x=0, x=1 and x=35.
Ivanshal [37]
Assuming the equation is x^2 + y^2 = 1, then that's a circle, with radius 1, centered on the origin [0,0].

So there are two tangents at x = 0. They are y = 1, and y = -1 (horizontal lines).

There is one tangent at x = 1. It is x = 1 (a vertical line).

There is no tangent at x = 35, because the original equation has no solution at x = 35.
4 0
3 years ago
In the 1990s the demand for personal computers in the home went up with household income. For a given community in the 1990s, th
WITCHER [35]

Answer:

a) 0.5198 computers per household

b) 0.01153 computers

Step-by-step explanation:

Given:

number of computers in a home,

q = 0.3458 ln x - 3.045 ;   10,000 ≤ x ≤ 125,000

here x is mean household income

mean income = $30,000

increasing rate, \frac{dx}{dt} = $1,000

Now,

a) computers per household are

since,

mean income of  $30,000 lies in the range of 10,000 ≤ x ≤ 125,000

thus,

q = 0.3458 ln(30,000) - 3.045

or

q = 0.5198 computers per household

b) Rate of increase in computers i.e \frac{dq}{dt}

\frac{dq}{dt} = \frac{d(0.3458 ln x - 3.045)}{dt}

or

\frac{dq}{dt}=0.3458\times(\frac{1}{x})\frac{dx}{dt} - 0

on substituting the values, we get

\frac{dq}{dt}=0.3458\times(\frac{1}{30,000})\times1,000

or

= 0.01153 computers

6 0
3 years ago
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