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muminat
3 years ago
10

Which symbol correctly compared these two fractions

Mathematics
1 answer:
rodikova [14]3 years ago
4 0

Answer:

B

Step-by-step explanation:

> = greater than

= this means equal to

< = less than

\frac{8}{13} =\frac{8}{13}

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A particle moves along the curve y=7 x 2+4y=7 x 2+4 in such a way that its xx-coordinate is changing at a rate of −5−5 centimete
ArbitrLikvidat [17]

Answer:

The y-coordinate is changing by the rate of -70 cm per sec.

Step-by-step explanation:

Given equation,

y = 7x^2 + 4

Differentiating with respect to time (t),

\frac{dy}{dt}=14x \frac{dx}{dt}

We have,

\frac{dx}{dt}=-5\text{ cm per sec}, x = 1

\frac{dy}{dt} = 14(1)(-5)=-70\text{ cm per sec}

7 0
3 years ago
Determine whether the equation has no solution, one solution, or infinitely many solutions. 7(12 – 2 x ) = 4(3 – 3 x )
Alex777 [14]

Answer: The equation has one solution

Step-by-step explanation: First , open the brackets

7( 12 - 2x ) = 4 (3 - 3x )

84 - 14x = 12 - 12x

collect the like terms

84 - 12 = - 12x + 14x

72 = 2x

Therefore divide through by 2

x = 72/2

x = 36

Therefore, the equation has only one solution

3 0
3 years ago
Is there a square made up of square tiles that has only 75 squares overall? What about one that has 120 squares overall? If it i
Mama L [17]

Answer:

See below.

Step-by-step explanation:

There are two essential pieces of information missing in the question, namely:

(1) "can the the square tiles be of varying size?"

(2) "what are the number constraints on the size of a tile?" (integers only? fractional lengths? Any real number?)

Without knowing the above, I can only speculate:

If the answer is "no, the square tiles are all same size, and must be of integer size" then the answer to your question will be "no, neither with 75 nor with 120 tiles a square of integer size can be covered" This is because the numbers 75 and 120 are not perfect squares (unlike, say, 121), which is easy to check.

If the answer is "yes, the square tiles can vary in size, but must be integers" then the answer to your question will be "yes" at least in the case of the 75 tiles...I could come up with an arrangement to solve that and am positive there is one for 120. I am not going into the detail here not knowing what is really required.

If the answer is "tile size can be any real number" then the answer to your question will be "yes" (albeit, there may never be possible to construct).

Let me know if you have questions.

7 0
3 years ago
How to set up and solve an equation to find the vertical line that bisects the region?
Inessa [10]

Answer: check explanation

Step-by-step explanation:

To find a general method on how to solve an equation to find the vertical line that bisects the region we will have to find the area between the curve. For instance, if we are to find a number A and the line y=A divides the region between the curved of y= X^2, say y= 4.

In the example given above, there is a need to solve for A. We will the take A and zero as the upper and lower limit boundaries respectively. That's ∫|(√4 - ydy)| = 16/3.

Then, 16/3 = 2 ∫(A - x^2) provided that the upper and lower limit boundaries are √b and 0 respectively. And, this is the case because because the function is even.

Therefore, A = 14/3.

The upper limit on the integral is where the curve meets the line y=A.

4 0
3 years ago
x=a+b+4 and a is inversely proportional to y²; b is inverseley propotional to 1/y. when y = 2 x = 18 and when y = 1 x = -3. find
AfilCa [17]

Answer:

When y = 4, x is equal to 39

Step-by-step explanation:

The given parameters are;

x = a + b + 4

a ∝ 1/y²

b ∝ 1/(1/y) = y

When y = 2, x = 18

When y = 1, x = -3

Therefore, we have;

a·y² = j

a = j/y²

b ∝ 1/(1/y)

∴ b ∝ y

b = k·y

When y = 2, x = 18, we have;

a = j/y² = j/2² = j/4

b = k·y = k·2

x = a + b + 4

∴ 18 = j/4 + k·2 + 4...(1)

When y = 1, x = -3, we have;

a = j/y² = j/1² = j

b = k·y = k·1 = k

x = a + b + 4

∴ -3 = j + k + 4...(2)

Making 'j', the subject of equation (1) and (2) gives;

From equation (1), we have;

18 = j/4 + k·2 + 4

∴ j = (18 - 4 - k·2) × 4 = 56 - 8·k

From equation (2), we have;

-3 = j + k + 4

∴ j = -3 - 4 - k = -7 - k

Equating the two values of 'j', gives;

56 - 8·k = -7 - k

56 + 7 = 8·k - k

63 = 7·k

k = 63/7 = 9

k = 9

From equation (2), we get;

-3 = j + k + 4

k = 9

∴ -3 = j + 9 + 4

j = -3 - 9 - 4 = -16

j = -16

When y = 4, we get;

x = a + b + 4

a = j/y²

b = k·y

∴ x = j/y² + k·y + 4

Plugging in the values of 'j', and 'k' and y = 4, gives;

x = (-16)/y² + 9·y + 4

∴ x = (-16)/4² + 9 × 4 + 4 = 39

x = 39

Therefore;

When y = 4, x = 39.

3 0
3 years ago
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