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Mila [183]
3 years ago
9

your class is going on a field trip. twenty-four have turned in their permission slip so far. this is 80% of the students in the

class. how many students are in the class
Mathematics
1 answer:
Mumz [18]3 years ago
8 0
There should be 30 students in the class as a total.
You might be interested in
Given m
Lilit [14]

Answer:

x = 41°, y = 139°

Step-by-step explanation:

The given parameters are;

Line <em>m</em> is parallel to line <em>n</em> and lines <em>m</em> and <em>n</em> have a common transversal

The corresponding angles formed by the common transversal to the two parallel lines are 41° on line <em>m</em> and <em>x°</em> on line <em>n</em>

Therefore, x° = 41° by corresponding angles formed between on two parallel lines by a common transversal are equal

x° and y° are linear pair angles and they are, supplementary

∴ x° + y° = 180°

∴ x° + y° = 41° + y° = 180°

y° = 180° - 41° = 139°

y° = 139°.

3 0
3 years ago
URGENT!!! please help me with these two problems, and solve with statement-reasoning. Thank you!!!
iren [92.7K]

Answer:

See below.

Step-by-step explanation:

By definition the median of the triangle bisects the  base of the isosceles triangle.

We need to prove that the 2 triangles formed by the median are congruent.

If the 2 triangles are ABD  and ACD  where BD is the median and < ABC is the angle from which BD is drawn.

BD = BD  ( the common side)

AD = DC ( because BD is the median).

AB =  AC ( because ABC is  an isosceles triangle).

So Triangles ABD and ACD are congruent by SSS.

Therefore  m < ABD = m <  CBD,  so BD is the bisector of < ABC.

To prove BD is also the altitude:

Triangles ABD and CBD are congruent as we have just proven. Therefore the

of measure of the base angle ABD = m < CBD . Also they are adjacent angles ( on the same line) so they add up to 180.

Therefore angles ABD and CBD are both  right angles and BD is the altitude of triangle ABC.

8 0
3 years ago
What is the order of the numbers from least to greatest
Mrrafil [7]

Answer:

-3.7, -1.9, -1/2, 1 1/3, 1.6, 2 1/3

Step-by-step explanation:

I believe this is correct, if not feel free to let me know and I will fix it. I'm sorry in advance if this answer is wrong.

4 0
3 years ago
Read 2 more answers
For the function defined by f(t)=2-t, 0≤t&lt;1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
3 years ago
You have $83 in your bank account each week you plan to deposit eight dollars from your allowance and $15 from your paycheck the
docker41 [41]
We know that you have 83$ in your bank account.

Okay so what is 15+8=23 So that means you get 23$ per week.

Now what is 83+23=106
106+23=129
129+23=152
152+23=175
Now that was four 23's


That means it would take 4 weeks for you to have 175 dollars in your bank account.
Hope this helped.
7 0
3 years ago
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