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2 years ago

8 grams = milligrams

Mathematics

2 answers:

saw5 [17]2 years ago

5 0

Answer: 8 grams is 8000 milligrams

Step-by-step explanation: 1 gram is equal to 1000 milligrams

Digiron [165]2 years ago

4 0

Answer:

8,000 miligrams

Step-by-step explanation:

There's 1000 milligrams in a gram so you would have to do multiplication.

which if you do..

8x1000

you should get...

8000 milligrams

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Help me please thx so much have a good one

Crazy boy [7]

Associative property of multiplication

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2 years ago

PLEASE HELP!!!!!!!

lozanna [386]

The correct answer is: C) The base of the cone and the top of the cylinder have the same area. The cone has the smallest volume of the 2 figures. This is because the formula for the cylinder is b x h, the formula for the cone is 1/3(b x h) so if they have same height and base area cylinder would have larger volume because, for the cylinder, a formula is one-third of b x h. Hope I helped!! : )

hope this helps you

4 0

2 years ago

Quadrilateral LMNO is similar to quadrilateral PQRS. Find the measure of side QR.

dalvyx [7]

Answer:

The measure of the side $QR$ is 8.8.

Step-by-step explanation:

Since $LMNO \sim PQRS$ , then $SP \propto OL$ , $RS \propto ON$ , $RQ \propto MN$ and $QP \propto LM$ . From figure we have the following relationship:

$k = \frac{OL}{SP} = \frac{ON}{RS} = \frac{MN}{QR} = \frac{ML}{QP}$ (1)

Where $k$ is the proportionality ratio.

If we know that $SP = 13, OL = 55, MN = 37$ , then the measure of side $QR$ is:

$k = \frac{OL}{SP}$ (1b)

$k = \frac{55}{13}$

$k = \frac{MN}{QR}$

$QR = \frac{MN}{k}$ (1c)

$QR = \frac{37}{\frac{55}{13} }$

$QR = 8.745$

The measure of the side $QR$ is 8.8.

4 0

2 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

2 years ago

What are to shapes that have to parallel sides

faltersainse [42]

The parallelogram, rectangle, and square all have two pairs of parallel sides

4 0

2 years ago

Read 2 more answers