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bagirrra123 [75]
2 years ago
11

How many times larger is the diameter than the radius of a circle?

Mathematics
2 answers:
UNO [17]2 years ago
6 0
  • The diameter is always twice the radius, so either form of the equation works.

<h2>HOPE IT HELPS❤</h2>
Effectus [21]2 years ago
4 0

Answer:

The diameter is always twice the radius.

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Sewing patterns recommended buying extra fabric in case of mistakes. Darlene bought 7 yards of fabric, plus 20% extra fabric. Ch
Debora [2.8K]

Answer:

Darlene bought more fabric

Darlene bought 3.4 yards more

Step-by-step explanation:

Darlene = 7 yards + 20% extra

= 7 yards + (20% of 7 yards)

= 7 + (20% × 7)

= 7 + (0.2 × 7)

= 7 + 1.4

= 8.4 yards

Chris = 4 yards + 25% of 4 yards

= 4 yards + (25% of 4 yards)

= 4 + (25% × 4)

= 4 + (0.25 × 4)

= 4 + 1

= 5 yards

Darlene bought more fabric

How much more?

Darlene - Chris

= 8.4 yards - 5 yards

= 3.4 yards

Darlene bought 3.4 yards more

6 0
3 years ago
For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec
telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

4sin(x^2) = 0

sin(x^2) = 0

x^2 = n\pi where n = 0, 1, 2,3, 4, 5,...

x = \sqrt(n\pi)

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0

xcos(x^2) = 0

x = 0 or

cos(x^2) = 0

x^2 = \frac{\pi}{2} + n\pi where n = 0, 1, 2, 3

x = \sqrt{\pi(n+1/2)}

Since we are restricting x between 0 and 3 we can stop at n =  2

So our critical points are at

x = 1.25, y = f(1.25) = 4sin(1.25^2) = 4

x = 2.17 , y = f(2.17) = 4sin(2.17^2) = -4

x = 2.8, y = f(2.8) = 4sin(2.8^2) = 4

For the inflection point, we can take the 2nd derivative and set it to 0

f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0

cos(x^2) = 2x^2sin(x^2)

tan(x^2) = \frac{1}{2x^2}

We can solve this numerically to get the inflection points are at

x = 0.81, y = f(0.81) = 4sin(0.81^2) = 2.44

x = 1.81, y = f(1.81) = 4sin(1.81^2) = -0.54

x = 2.52, y = f(2.52) = 4sin(2.52^2) = 0.27

3 0
3 years ago
IMAGE BELOW I WILL GIVE BRAINLIEST
diamong [38]

Answer:

for L 2, 2 and for LM it is -8,8

Step-by-step explanation:

easy

7 0
2 years ago
Please help it is TIMED.
lions [1.4K]

Answer:

message it will be easy er to explain

Step-by-step explanation:


4 0
3 years ago
Read 2 more answers
4(4-w)=3(2w+2)<br>find w​
Travka [436]

Answer:

1

Step-by-step explanation:

4(4-w) = 3(2w+2)

Use distributive property:

16 - 4w = 6w + 6

    +4w  +4w

------------------------

     16 = 10w + 6

    -6              -6

-------------------------

        10 = 10w

        ----    -----

        10      10

         1 = w

Hope this helped.

7 0
3 years ago
Read 2 more answers
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