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Nataliya [291]
3 years ago
9

the distance between two lakes is 300 miles. on a map, the measured distance between the lakes is three-fourths of costs $24.00

for 6 quarts?
Mathematics
1 answer:
Ghella [55]3 years ago
3 0

1 inch = 400 miles The scale gives the relationship between the measurements of the model and the measurements of the real object. Hope this helps.

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Can someone please help?
Ber [7]

Answer:

  (B)  26°

Step-by-step explanation:

The angle at A made by the radius and the tangent is 90°. The angle at O is the same as arc AB, 64°. The acute angles in a right triangle are complementary, so the angle at C is the complement of 64°.

∠ACB = 90° -64°

∠ACB = 26°

5 0
3 years ago
I need help with this asappp
Rus_ich [418]
The answer would be 45. Thank you very much and good luck on there!
6 0
3 years ago
Read 2 more answers
Consider an experiment with a deck of 52 playing cards, in which there are 13 cards in each suit, two suits are black, and two s
lakkis [162]

Answer:

<u>P (E1) = 1/2 or 50%</u>

<u>P (E2) = 3/13 or 23%</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Number of playing cards = 52

Number of suits = 4

Number of cards per suit = 13

Number of black suits = 2

Number of red suits = 2

2.  Suppose E1 = the outcome is a red card and E2 = the outcome is a face card (K, Q, J). Determine P(E1 or E2).

P (E1) = Number of red cards/Total of playing cards

P (E1) = 26/52 = 1/2 = 50%

P (E2) = Number of face cards/Total of playing cards

P (E2) = 12/52 = 3/13 = 23%

8 0
3 years ago
Read 3 more answers
Which system of linear inequalities is represented by the graph?
masha68 [24]
For the red line, we have:
 
gradient (m) = 1/3
y-intercept = 3

Equation of straight line is given by y=mx+c, where m is the gradient and c is where the line intercept the y-axis

So we have, y= \frac{1}{3}x+3

The shaded part is above the line, hence the first inequality is
y \geq \frac{1}{3}x+3

The black line, we have
gradient, (m) = 3
y-intercept = -2

y=mx+c
y=3x-2, rearranging gives
3x-y=2

The shaded region is to the right of the line, hence the inequality is
3x-y\ \textgreater \ 2

Note: The first option should be the correct answer but the inequality sign for <span>y > 1/3x + 3 should be </span>\geq 



6 0
3 years ago
Read 2 more answers
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
3 years ago
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