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serious [3.7K]
4 years ago
11

What is the prime factorization of 74

Mathematics
1 answer:
mash [69]4 years ago
3 0
74 is a composite number. 74=1*74 or 2*37. Factors of 74: 1,2,37,74. Prime factorization: 74=2*37.
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Hello, this is math and I need help ❤️ pls and ty ONLY SERIOUS ANSWERS
Greeley [361]

The bicycle with a 30-inches diameter rolls approximately 28.8 more than a tire with the diameter with 21-inches.

<u><em>How to solve:</em></u>

  • First use the formula: C = 3.14d
  • Replace the d value with 30 and solve the equation:

C = 3.14 x 30    ⇒   94.2477796077

  • Then solve the equation again but 21 as the d value:

C = 3.14 x 21     ⇒   65.9734457254

  • Then subtract 65.9734457254 from 94.2477796077 to get the approx. answer of 28.8

3 0
3 years ago
What is the distance between points (1, 4) and (9, 16)? round to the nearest tenth.
11Alexandr11 [23.1K]

Answer:

14.4 to nearest tenth.

Step-by-step explanation:

By the Pythagoras theorem:-

Distance = √ [ (y2-y1)^2 + (x2-x1)^2 ]   where  the 2 points are (x1,y1) and (x2,y2)

= √ [ (16-4)^2 + (9-1)^2]

= √ 208

= 14.4

6 0
3 years ago
Help me please I donr understand this ​
Serggg [28]
For the function w,w(9)

8 0
3 years ago
Read 2 more answers
I will give brainliest if anyone knows this
mezya [45]

Answer:

7 4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.
sergeinik [125]

A generic point on the graph of the curve has coordinates

(x, 4x^2+1)

The derivative gives us the slope of the tangent line at a given point:

f(x) = 4x^2+1 \implies f'(x) = 8x

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through (k, 4k^2+1) and have slope 8k

So, we can write its equation using the point-slope formula: a line with slope m passing through (x_0, y_0) has equation

y-y_0 = m(x-x_0)

In this case, (x_0, y_0)=(k, 4k^2+1) and m=8k, so the equation becomes

y-4k^2-1 = 8k(x-k)

We can rewrite the equation as follows:

y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1

We know that this function must give 0 when evaluated at x=0:

f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}

This equation has no real solution, so the problem looks impossible.

5 0
3 years ago
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