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Stels [109]
3 years ago
6

43cm =how many meters

Mathematics
2 answers:
Katena32 [7]3 years ago
7 0
Answer:  43 cm = 0.43 meters   
notsponge [240]3 years ago
3 0
1 meter is equal to 100cm,
therefore, 43cm is equal to 0.43m

Hope this helps :) 
You might be interested in
Write an algebraic rule for the trend line. Report your rule in slope intercept form.
Tju [1.3M]

Answer:

y = 5x + 20

Step-by-step explanation:

Here, we want to find the slope intercept form equation of the trend line

The general form is:

y =

mx + b

b is slope and m

is the y-intercept

b = 20

To get the slope, we select any two points and use the slope formula

we have the points;

(10,70) and (5,45)

The slope formula is;

m = (y2-y1)/(x2-x1) = (45-70)/(5-10) = -25/-5 = 5

The equation of the line is thus;

y = 5x + 20

4 0
3 years ago
A company makes 140 bags. 39 of the bags have buttons but no zips. 21 of the bags have zips but no buttons. 23 of the bags have
sleet_krkn [62]

Subtraction is a mathematical operation that reflects the removal of things from a collection. The number of bags that zips on them is 78.

<h3>What is subtraction?</h3>

Subtraction is a mathematical operation that reflects the removal of things from a collection. The negative symbol represents subtraction.

Given the number of bags that the company makes is 140. 39 of the bags have buttons but no zips. 21 of the bags have zips but no buttons. 23 of the bags have neither zips nor buttons.

Bags that have both zip and button

= Total number of bags - Number of bags with zips but no buttons - Number of bags with buttons but no zips - Number of bags with nor zips nor the buttons

= 140 - 21 - 39 - 23

= 57

Now, the number of bags that has zips will be the bags that only have zips and the number of bags that has zips and buttons. Therefore, the number of bags with zips is,

n = Bags that have both zip and button + Number of bags with zips but no buttons

   = 37 + 21

   = 78

Hence, the number of bags that zips on them is 78.

Learn more about Subtraction:

brainly.com/question/1927340

#SPJ1

4 0
1 year ago
Read 2 more answers
The difference in the x-coordinates of two points is 3, and the difference in the y-coordinates of the two points is 6. What is
oksano4ka [1.4K]
It Does Not Matter Where You Put The Line, As The Slope Stays The Same. So, We Can Say That One Point Is (3,0)
(3,0) and (6,6)
So, The Slope Is 2.
6 0
3 years ago
Read 2 more answers
Brooke paints the outsides of the square walls and triangular ceilings of her treehouse. What area does she paint? Enter your an
andrew-mc [135]

Answer:

216

Step-by-step explanation:

The formula of finding the surface is length * width * height or LxWxH.

Since everything is 6 you just multiply 6*6*6 which gives you 36*6 which gets you 216. You don't need the top 6 just to find the surface she will paint.

Hope this helps!

7 0
3 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
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