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3 years ago

Convert: 42 centigrams 6 milligrams = 1 milligrams

Mathematics

1 answer:

Naddik [55]3 years ago

6 0

Answer:

centigram is 'bigger.'

Well, if you look on the metric scale it is apparent that 1 mg = 0.1 (or 1/10) cg suggesting that a milligram is the smaller value.

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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

Please help please please help 16 points

asambeis [7]

Answer:

1.6 = x axis || 2.5 = y axis

Step-by-step explanation:

each line is 0.5

4 0

2 years ago

Please help me asap.

Makovka662 [10]

The first one is the answer you're looking for

6 0

3 years ago

Please help! Vectors and angles

BaLLatris [955]

The velocity of the ship = 22 ∠157° = (22 cos 157°) i+ (22 sin 157°) j

The velocity of current = 5 ∠213° = (5 cos 213°) i+ (5 sin 213°) j

So, the resultant velocity = 22 ∠157° + 5 ∠213°
= (22 cos 157°) i+ (22 sin 157°) j + (5 cos 213°) i+ (5 sin 213°) j
= (22 cos 157° + 5 cos 213°) i + (22 sin 157° + 5 sin 213°) j
= -24.444 i + 5.873 j
= 25.14 ∠166.5°

The correct answer is option (1)
<span>1) 25 knots at 166.5 degree</span>

6 0

3 years ago

Read 2 more answers

How to do number 2 and 5 !!! Please someone help

Brums [2.3K]

2a) there is a right angle on T so 180-90-37=53
2b) do pythogrean theorem 22sq+ 12sq= 628 sq rt of 628 is 25.06 which is 25
5a) 5*8=10x so X=4
5b) 47*2=94 -57 = 37

4 0

3 years ago