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3 years ago

5 points free, good job

Mathematics

1 answer:

Vilka [71]3 years ago

4 0

Answer:

ok, what's the question?

Step-by-step explanation:

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Tju [1.3M]

Answer:

Step-by-step explanation:

5 0

2 years ago

A commercial package contains thirty-six 200-mg tablets of ibuprofen. How many kilograms of ibuprofen were

PIT_PIT [208]

Answer: 0.0072 kilogram

Step-by-step explanation:

Given : The mass of one tablet of ibuprofen =200 mg

Then, the mass of 36 tablets of ibuprofen = $36\times200$ mg

i.e. the mass of 36 tablets of ibuprofen = $7200$ mg

We know that 1 kilogram = 1000 grams

and 1 gram = 1000 milligram

Then, 1 kilogram = $1000,000$ milligrams

Then, $\text{1 mg}=\dfrac{1}{1000,000}\text{ kg}$

Now, $\text{7200 mg}=\dfrac{7200}{1000,000}=0.0072\text{ kg}$

Hence, A commercial package contains 0.0072 kilogram of ibuprofen.

3 0

3 years ago

What is the answer I need help

Softa [21]

Answer: A 8040 is exactly divisible by 67 the quotient is 120

3 0

2 years ago

4/8 x 6 answer first gets 10000 points

natulia [17]

Answer:

Step-by-step explanation: now weres the 10000 points you scammer

3 0

3 years ago

Read 2 more answers

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0

3 years ago