All categories

2 years ago

What is the sum of the first five prime numbers? A 13 B 26 C 28 D 29

Mathematics

2 answers:

Sholpan [36]2 years ago

8 0

28 or C

Prime numbers are numbers that <em>cannot</em> be made by multiplying other numbers.

The first five prime numbers are 2, 3, 5, 7, and 11.

Add them together to get 28.

2+3+5+7+11

= 28

slavikrds [6]2 years ago

6 0

The first prime numbers are 1 and 3 ,5,7,9 Added together the total is 25

You might be interested in

What is the value of the x-coordinate that is 6 units to the right of (3.25,-4)?

hammer [34]

9.25 hope I have a nice day and stuff I have to write 20 words so

8 0

3 years ago

A. 9+ 2 < 17 or 7− 4 < −9

Ira Lisetskai [31]

Answer:9+2<17

Step-by-step explanation:

4 0

3 years ago

the radius of a circle is 2 cm. A point 8cm away from the center of the circle. What is the length of the tangent drawn to the c

SIZIF [17.4K]

Answer:

7.75 cm

Step-by-step explanation:

Let a circle having centre O and AB is a tangent of the circle and OB is the radius of the circle = 2 cm. And join AO such that AO = 8 cm.

$AB=\sqrt{(AO)^{2}- (OB)^{2}} cm$

$AB=\sqrt {8^{2}-2^{2}}=\sqrt{60}=7.745966692 cm\approx 7.75 cm$

3 0

3 years ago

A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 3 feet, 55 feet awa

MariettaO [177]

Answer:

The maximum height of the second pluck is greater than that of the first pluck, hence the second pluck travels further

Also the distance of the maximum height of first pluck from the player is less than the distance of the second pluck from the player hence the second pluck travels further

Step-by-step explanation:

From the question we are told that

The maximum height of the first pluck is $h_1 = 3 \ ft$

The height of the second height is mathematically represented as

$$y=x\left(0.15-0.001x\right)$

=> $y=0.15x -0.001x^2$

Generally at maximum height $y' = 0$

$y'=0.15 -0.002 x = 0$

=> $x = 75 \ ft$

Here 75 ft is the horizontal distance the second pluck traveled at maximum height

So the maximum height of the second pluck is mathematically represented as

$y=0.15(75) -0.001(75)^2$

=> $y= 5.625 \ ft$

So comparing the maximum height of the first and the second pluck we see that the maximum height of the second pluck is greater than that of the first pluck, hence the second pluck travels father

3 0

3 years ago

Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Co

Volgvan

Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like $v_{1},v_{2},v_{3}, ...$ we call them linearly dependent if we have scalars $a_{1},a_{2},a_{3},...$ as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

$a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0$

When all scalar coefficients are equal to zero, we can call them linearly independent

2) Now let's examine the Matrix given:

$\begin{bmatrix}1 &-2 &2 &3 \\ -2 & 4 & -4 &3 \\ 0&1 &-1 & 4\end{bmatrix}$

So each column of this Matrix is a vector. So we can write them as:

$\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle$ Or

Now let's rewrite it as a system of equations:

$a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

$\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ -2 &4 &-4 &3 \\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )R_{1}\times2 +R_{2}\rightarrow R_{2}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &0 &9 &0\\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\ R_{2}\Leftrightarrow R_{3}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &1 &-1 &4 \\ 0 &0 &9 &0 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )$ $\left\{\begin{matrix}1a_{1} &-2a_{2} &+2a_{3} &+3a_{4} &=0 \\ &1a_{2} &-1a_{3} &+4a_{4} &=0 \\ & & &9a_{4} &=0 \end{matrix}\right.\Rightarrow a_{1}=0, a_{2}=a_{3},a_{4}=0$

$S=\begin{bmatrix}0\\ a_{3}\\ a_{3}\\ 0\end{bmatrix}$

3 0

3 years ago