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Lerok [7]
3 years ago
13

Which of the following are true statements about David Hilbert?

Mathematics
1 answer:
qwelly [4]3 years ago
7 0

Answer:

this is your answer

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Please answer seriously will give brainliest
SashulF [63]

Answer:

Step-by-step explanation:

33)

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Secant - GE

Diameter -  FE

Radius -  CG, CF ,  CE

Point of tangency - D

34

Chord  - JL

Secant - MN

Diameter - JL

Radius - RK , JK , KL

Point of tangency - U

35)Chord  - PN

Secant -  LM

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7 0
2 years ago
5.which of the following ordered pairs is a solution of the given system of linear equations
liq [111]

Answer:

the answer is d

Step-by-step explanation:

answers are shown

4 0
3 years ago
Which systems of equations intersect at point A in this graph?
abruzzese [7]

Answer:

The point of intersection of the system of equations is:

(x, y) =  (-2, 1)

The correct system of equations intersect at point A in this graph will be:

\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}

Thus, the second option is correct.

Step-by-step explanation:

Given the point

  • A (-2, 1)

Let us check the system of equations to determine whether it intersect at point A in this graph.

Given the system of equations

\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}

Arrange equation variable for elimination

\begin{bmatrix}y-4x=9\\ y+3x=-5\end{bmatrix}

so

y+3x=-5

-

\underline{y-4x=9}

7x=-14

so the system of equations becomes

\begin{bmatrix}y-4x=9\\ 7x=-14\end{bmatrix}

Solve 7x = -14 for x

7x=-14

Divide both sides by 7

\frac{7x}{7}=\frac{-14}{7}

Simplify

x = -2

For y - 4x = 9 plug in x = 2

y-4\left(-2\right)=9

y+4\cdot \:2=9

y+8=9

Subtract 8 from both sides

y+8-8=9-8

Simplify

y = 1

Thus, the solution to the system of equations is:

(x, y) = (-2, 1)

From the attached graph, it is also clear that the system of equations intersects at point x = -2, and y = 1.

In other words, the point of intersection of the system of equations is:

(x, y) =  (-2, 1)

Therefore, the correct system of equations intersect at point A in this graph will be:

\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}

Thus, the second option is correct.

3 0
2 years ago
Jonah has a large collection of marbles he notices that if he borrows five marbles from a from a friend he can arrange the marbl
olga nikolaevna [1]

Answer:

The remainder is 8

Step-by-step explanation:

Let Jonah's Marble=x

If he arranges x marbles into y rows of 13 each, and there are remainders(R)

Then:

x/13=y+(R/13).....(I)

If he borrows 5 marbles from a friend, there will be no remainder. However, the number of rows y, will be increased by 1

New Total Marbles=x+5

(x+5)/13=y+1.....(ii)

x+5=13(y+1)

x=13y+13-5

x=13y+8

From (I)

x=13y+R

Comparing the values of x derived from (I) and (ii)

13y+R=13y+8

Therefore the Remainder, R= 8

5 0
3 years ago
Explain how to multiply the following whole numbers 21 x 14
Lesechka [4]

Answer:

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Step-by-step explanation:

Given

21\:\times \:14

Line up the numbers

\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

Multiply the bold numbers:    1×4=4

\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}

Multiply the bold numbers:    2×4=8

\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the top number by the bolded digit of the bottom number

\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the bold numbers:    1×1=1

\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}

Multiply the bold numbers:    2×1=2

\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}

Add the rows to get the answer. For simplicity, fill in trailing zeros.

\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}

adding portion

\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}

Add the digits of the right-most column: 4+0=4

\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}

Add the digits of the right-most column: 8+1=9

\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}

Add the digits of the right-most column: 0+2=2

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Therefore,

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

6 0
3 years ago
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