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Marrrta [24]
3 years ago
13

Jake tossed a paper cup 50 times and recorded how it landed. The table shows the results:

Mathematics
1 answer:
Dmitry [639]3 years ago
8 0
Hello!

Here are 6 steps to Help You Out!

1. 3 divided by 50 = .06

2. Then if you like you can turn it in to a percent not sure how the answer is supposed to look for example:

<span><span>3. 3/50</span>=<span>6/100</span>=6%

</span>4. 7 divided by 50 - .14

5. so 0.06 for the first one

6. And 40 divided by 50 = .8

====>So as we can see, o<span>pen side up .06 closed side up .14 and  on its side .8 or open 6% closed 14% and side 80% chance.<====

Hope this Helps! Have A WONDERFUL Day! :)</span>
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B<br>Find all the missing angles - a,b,c, and d​
mart [117]

a=65° d=75° b=25° c=25°

4 0
2 years ago
Read 2 more answers
In the figure, triangle CAE is an enlargement of triangle CBD with scale factor of 4/3. The area of the smaller triangle is 9cm2
balandron [24]

Answer:

16 cm^2

Step-by-step explanation:

Given

\triangle CAE -- Bigger Triangle

\triangle CBD -- Smaller Triangle

k = \frac{4}{3} --- Scale factor

Area of CBD = 9

Required

Determine the area of CAE

The area of triangle CBD is:

A_1 = \frac{1}{2}bh

\frac{1}{2}bh = 9

The area of CAE is:

A_2 = \frac{1}{2}BH

Where:

B = \frac{4}{3}b and

H = \frac{4}{3}h

The above values is the dimension of the larger triangle (after dilation).

So, we have:

A_2 = \frac{1}{2}*\frac{4}{3}b * \frac{4}{3} * h

A_2 = \frac{1}{2}*\frac{4}{3} * \frac{4}{3} *b* h

A_2 = \frac{1}{2}*\frac{16}{9}  *b* h

Re-order

A_2 = \frac{16}{9}*\frac{1}{2}* b* h

A_2 = \frac{16}{9}*\frac{1}{2}bh

Recall that:

\frac{1}{2}bh = 9

A_2 = \frac{16}{9}*9

A_2 = 16

Hence, the area is 16 cm^2

3 0
3 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
2 years ago
Which ordered pairs lie on the graph of the exponential function f(x)=−3^2x+5 ? Select each correct answer. ​ (−2,  76) ​ ​ ​​ (
In-s [12.5K]

Plugging in values, we get that (1,-4) works.

f(1)=-3^{2*1}+5=-3^2+5=-9+5=-4


3 0
3 years ago
Read 2 more answers
Please help sorry if it’s a little blurry! I’ll mark brainliest if you tell me why you got that answer.
sergij07 [2.7K]

Answer:

Hi! Saw your comment saying you got it, congrats! I hope you are having an amazing day/night and I wish you luck with the rest of your studies! I hope you don't mind that I use this for some points! <3

3 0
3 years ago
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