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3 years ago

X over 13 is greater than or equal to 4... answer?

Mathematics

2 answers:

Olin [163]3 years ago

5 0

x ≥ 4 is x/13 times 13 ≥ 4 times 13
13

That means that x ≥ 52.

BaLLatris [955]3 years ago

4 0

In mathematical expression, You can write it as:
x/13 ≥ 4

Multiply both sides by 13,
x/13 * 13 ≥ 4 * 13
x ≥ 52

In short, Your Answer would be x ≥ 52

Hope this helps!

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balandron [24]

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3 years ago

A school has 300 students and 30 teachers. What is the ratio between the number of teachers and the number of students of the sc

Oduvanchick [21]

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2 years ago

Can someone help me with this problem

Illusion [34]

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$\sqrt{3} = 1.73$

7 0

3 years ago

Manufacture wants to enlarger it's floor area 1.5 times that of the current facility. The current facility is 260 ft by 140 ft.

Lady bird [3.3K]

Answer:

New dimensions of the floor is approximately 301.25 ft by 181.25 ft

Step-by-step explanation:

The question is incomplete. The complete question should be:

Manufacture wants to enlarger it's floor area 1.5 times that of the current facility. The current facility is 260 ft by 140 ft. The manufacture wants to increase each dimension the same amount. Write the dimensions of the new floor.

Given:

Length of floor = 260 ft

Width of floor = 140 ft

The floor area is increased 1.5 times.

To find the new dimensions of the floor.

Solution:

Original area of the floor = $length\times width= 260\times 140=36400\ ft^2$

New area = $1.5\times Original\ Area = 1.5\times 36,400=54,600\ ft^2$

Let the length and width be increased by $x$ ft.

Thus, new length = $(260+x)\ ft$

New width = $(140+x)\ ft$

Area of the new floor can be given as:

⇒ $new\ length\times new\ width$

⇒ $(260+x)(140+x)$

Multiplying using distribution.

⇒ $x^2+260x+140x+36400$

⇒ $x^2+400x+36400$

Thus we can equate this with new area to get the equation to find $x$

$x^2+400x+36400=54600$

subtracting both sides by 54600.

$x^2+400x+36400-54600=54600-54600$

$x^2+400x+18200=0$

Using quadratic formula:

For a quadratic equation $ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

For the equation $x^2+400x-18200=0$

$x=\frac{-400\pm\sqrt{(400)^2-4(1)(-18200)}}{2(1)}$

$x=\frac{-400\pm\sqrt{232800)}}{2}$

$x=\frac{-400\pm482.49}{2}$

$x=\frac{-400+482.49}{2} \ and\ x= \frac{-400-482.49}{2}$

∴ $x\approx 41.25 \ and\ x\approx-441.25$

Since length is being increased, so we take $x\approx41.25$

New dimensions are:

New length $\approx 260\ ft + 41.25\ ft =301.25\ ft$

New width $\approx 140\ ft + 41.25\ ft =181.25\ ft$

3 0

3 years ago

I need help to solve this and also Hearn will the bacteria be over 100,000

lapo4ka [179]

Hello! Let's look at the two parts of this question.

Complete the table:

In this case, you just substitute the value of "hour" into the equation, for the value of t. For example:

P(0) = 120 $(2)^{0}$

P(0) = 120 (1)

P(0) = 120

Therefore, the number of bacteria for hour 0 is 120.

You can do this for the next ones. Hour 1 = 240, hour 2 = 480, and so on. (In this case, you can keep multiplying by 2)

Estimate when there will be more than 100,000 bacteria:

Set the final value of P(t) = 100,000, then solve.

100,000 = 120 (2 $)^{t}$

833.33 = (2 $)^{t}$

t = $log_{2}833.33$

t = 9.702744108

So your answer would be around 9.7 years, or, around 10 years.

Hope this helps!

3 0

3 years ago