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Fynjy0 [20]
3 years ago
9

" alt="6 = \frac{a}{4} + 2" align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
natta225 [31]3 years ago
7 0
It is 1 Bc if u multiple 1 times 4 and then add 2 it will give you 6welcome
Olegator [25]3 years ago
4 0
A=16 because 16/4=4
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just olya [345]
I believe the answer is the fourth option.
7 0
2 years ago
Please Help Me :((.
Sidana [21]

Answer:

4.2 h

Step-by-step explanation:

12 * (50%-15%) = 4.2 h

8 0
2 years ago
Which action is not a step in using paper folding to find the midpoint of a line
dedylja [7]

Answer:

A. Draw a line from the segment to any point on the fold line.

Step-by-step explanation:

When using the paper folding method to find the midpoint of a line segment we take the following steps:

  1. Draw a line segment on the tracing paper
  2. Fold the tracing paper so that the endpoints lie on top of each other
  3. Mark the intersection of the fold and the segment with a point

These steps include statements B, C, and D. Drawing a line from the segment to any point on the fold line, which is statement A,  is not included in these steps because it is not needed.

Thus, choice A is not a step in using paper folding to find the midpoint of a line segment.

5 0
3 years ago
3. 18 is 40% of what number?​
Mariulka [41]

#Lets just write x as the nunber you asked

x × 40/100 = 18

x = 18 : 40/100

x = 18 × 100/40

y = 1800/40

y = 45

<h3>Answer : 45</h3>
8 0
2 years ago
Read 2 more answers
Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len
Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> )   ==>   d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em>   ==>   d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt

=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt

=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt

=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt

=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}

5 0
2 years ago
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