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wariber [46]
2 years ago
15

Match the shapes that have same preimeters

Mathematics
1 answer:
Alexus [3.1K]2 years ago
7 0

Answer:

sorry i can't help i don't understand the question

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Csc0sin20-sec0=cos20sec0​
Mkey [24]

Answer:0

Step-by-step explanation:

(0)0.342-0=0.9397(0)

0=0

sin 20= 0.342

cos 20=0.9397

3 0
2 years ago
Kendra is buying an outfit that is on sale for 20% off. The original price of the outfit is $109.
Misha Larkins [42]
It could be 64$ because i did division

4 0
2 years ago
Read 2 more answers
Determine whether the following are subspaces of R2×2 : (a) Thesetofall2×2diagonalmatrices (b) The set of all 2 × 2 triangular m
Mrrafil [7]

A vector space V is a subspace of a vector space W if

  • V is non-empty,
  • for any two vectors v_1,v_2\inV we have v_1+v_2\in V, and
  • for any scalar k and v\in V we have kv\in V.

It's easy to show the first condition is met by all the sets in parts (a-g).

(a) is a subspace of \Bbb R^{2\times2} because adding any 2x2 diagonal matrices together, or multiplying one by some scalar, gives another diagonal matrix.

(b) and (c) are also subspaces for the same reasons.

(d) is not a subspace because \Bbb R^{2\times2} because this set of matrices does not contain the zero matrix.

(e), however, is a subspace. Any linear combination of matrices in this set always yields a matrix with 0 in row 1, column 1 entry.

(f) is a subspace. A symmetric matrix is one of the form

\begin{bmatrix}a&b\\b&c\end{bmatrix}

Adding two symmetric matrices gives another symmetric matrix:

\begin{bmatrix}a_1&b_1\\b_1&c_1\end{bmatrix}+\begin{bmatrix}a_2&b_2\\b_2&c_2\end{bmatrix}=\begin{bmatrix}a_1+a_2&b_1+b_2\\b_1+b_2&c_1+c_2\end{bmatrix}

(g) is not a subspace. Consider the matrices

\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}

Both matrices have determinant 0, but their sum is the identity matrix with determinant 1.

4 0
3 years ago
A pebble is dropped into a deep well and 3s later the sound of a splash is heard as the pebble reaches the bottom of the well. T
ankoles [38]

Answer:

2.89\ \text{s}

0.11 seconds

37.4\ \text{m}

Step-by-step explanation:

t_1 = Time taken by the pebble to hit the water

t_2 = Time taken by the sound to travel to the observer

t_1+t_2 = Time taken by the observer to listen to the splash after stone is thrown = 3 s

t_1=3-t_2

a = g = Acceleration due to gravity = 9.81\ \text{m/s}^2

u_s = Speed of sound = 340 m/s

The distance traveled by the pebble

s=ut+\dfrac{1}{2}at_1^2\\\Rightarrow s=0+\dfrac{1}{2}\times9.81(3-t_2)^2\\\Rightarrow s=4.905(9+t_2^2-6t_2^2)

Distance traveled by the sound

s=u_st\\\Rightarrow s=340\times t_2

The distance traveled by the sound and the pebble is equal

4.905(9+t_2^2-6t_2)=340\times t_2\\\Rightarrow 9+t_2^2-6t_2=\dfrac{340}{4.905}t_2\\\Rightarrow 9+t_2^2-6t_2=69.32t_2\\\Rightarrow t_2^2-75.32t_2+9=0\\\Rightarrow t_2=\frac{-\left(-75.32\right)\pm \sqrt{\left(-75.32\right)^2-4\times \:1\times \:9}}{2\times \:1}\\\Rightarrow t_2=75.2,0.11

Since t_1+t_2=3 the value of t_2 cannot be greater than 3.

So, time taken by the sound to reach the observer is 0.11 seconds.

Time taken by the pebble to hit the water is t_1=3-t_2=3-0.11=2.89\ \text{s}

Depth of the well is s=340t_2=340\times 0.11=37.4\ \text{m}

8 0
2 years ago
Factor -4 out of -24a - 44
disa [49]

Answer:

After factoring -4

-4(6a+11)

7 0
2 years ago
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