Y=-7/5x-3 as a proper fraction

Mathematics

1 answer:

Ann [662]3 years ago

3 0

Answer:

5x-3

Step-by-step explanation:

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Pleassdee need help I'll brainlist

NikAS [45]

Answer:

A^2 means a*a

Step-by-step explanation:

A^2 = 5^2=5*5=25

3a^2=3*5^2=3*25=75

A^2+7=5^2+7=25+7=32

5 0

2 years ago

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Hayley begins solving this problem by combining like terms. 3x + 5x = 10 Which problem requires the same strategy (combining lik

Tamiku [17]

D) 3x - 2x = 10 is the answer to this equation

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3 years ago

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What is the solution to this system of equations?

Degger [83]

Answer:

Step-by-step explanation:

8 0

2 years ago

The length of a photograph is 3cm less than twice the width. The area is 54cm^2. Find the dimensions of the photograph.

REY [17]

W= width
l= length= 2w-3

Area= length * width
A= lw
substitute l=2w-3 for l
54= (2w-3)(w)
54= 2w^2 - 3w
subtract 54 from both sides
0= 2w^2 - 3w - 54
factor the polynomial
0= (2w + 9)(w - 6)

Set each set of parentheses equal
to 0 and solve for width.

0=2w+9
-9=2w
-4.5= w NEGATIVE VALUE

0=w-6
6 cm=w

-4.5 is not the answer because it has a negative value. You cannot have a negative side of a photograph. So substitute 6 in to find length.

length= 2w-3
l= 2(6)-3
l= 12-3
l= 9 cm

CHECK:
w= 6cm
l= 9cm
Area= length * width
54= 6*9
54= 54

ANSWER: width= 6cm; length= 9cm

Hope this helps! :)

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2 years ago

A drunkard executes a “random walk” in the following way: each minute he takes a step north or southm with probability 1/2 each,

valentina_108 [34]

Assume that a step to the north is in the positive direction and a step in the south is in the negative direction.

Then,

$E(X_i)=50\left(\frac{1}{2}\right)-50\left(\frac{1}{2}\right)\\ \\=25-25=0$

and

$\sigma^2(X_i)=E(X_i^2)-E(X_i)^2 \\ \\ =50^2\left(\frac{1}{2}\right)+50^2\left(\frac{1}{2}\right)-0^2=2500\left(\frac{1}{2}\right)+2500\left(\frac{1}{2}\right)-0 \\ \\ =2500 \\ \\ \Rightarrow\sigma(X_i)=50$

After 1 hour of independent steps,

$E(X)=\sum_{i=1}^{60}E(X_i)=60(0)=0$

and

$\sigma^2(X)=\sum_{i=1}^{60}\sigma^2(X_i) \\ \\ =60(50)=300 \\ \\ \Rightarrow\sigma(X)=\sqrt{300}=17.32$

Therefore, by Central Limit Theorem, the distribution of the random walk is normal with a mean of 0 cm and a standard deviation of 17.32 cm. He is expected to be somewhere around his starting point after 1 hour.

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3 years ago

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