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Diano4ka-milaya [45]
3 years ago
5

Which of the following could be a function with zeros of —3 and 2? А f(x) = (x - 3)(x + 2) B f(x) = (x = 3)(x - 2) © f(x) = (x+3

)(x - 2) D f(x) = (x+3)(x + 2)​
Mathematics
1 answer:
Zinaida [17]3 years ago
6 0

Answer:

C

Step-by-step explanation:

to find the zeros you look at what value you would fill in for x to make zero, so (x+3) would have a zero of -3 and (x-2) would have a zero of 2

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Morgarella [4.7K]

Hello there! The answer is the first option, 31/55.

To solve this, we don't even need to do any math. Note that fractions with the same numerator and denominator will be equal to 1.

Knowing this and looking at our second and fourth options, 55/55 x 111 and 31/31 x 111, these problems are the same as 1 x 111, which results in 111, which is not less than, leaving us with the third and first options.

The third option is 55/31, meaning we have more than a whole, so we are multiplying by a number greater than 1, making our answer over 111 and this option not correct.

This leaves the first option as your answer!

7 0
3 years ago
Based on the picture below. find the angle measure for lmn.
trasher [3.6K]

Answer:

Step-by-step explanation:

4x + 2x = 90

6x = 90

x = 15

4(15)= 60 for <LMN

2(15)= 30

5 0
3 years ago
8x83x8303x838x83x9993x93939x99393
Zolol [24]

Answer:

3.5778559e+25

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Help me on this question please
Anika [276]

Answer:

48

Step-by-step explanation:

The formula for finding the area of a rectangle is: l * w = a

So 12*4 = 48.

3 0
3 years ago
EXAMPLE 1 (a) Find the derivative of r(t) = (2 + t3)i + te−tj + sin(6t)k. (b) Find the unit tangent vector at the point t = 0. S
Tatiana [17]

The correct question is:

(a) Find the derivative of r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(b) Find the unit tangent vector at the point t = 0.

Answer:

The derivative of r(t) is 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is (j/2 + 3k)

Step-by-step explanation:

Given

r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(a) To find the derivative of r(t), we differentiate r(t) with respect to t.

So, the derivative

r'(t) = 3t²i +[e^(-t) - te^(-t)]j + 6cos(6t)k

= 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is obtained using the formula r'(0)/|r(0)|. r(0) is the value of r'(t) at t = 0, and |r(0)| is the modulus of r(0).

Now,

r'(0) = 3t²i + (1 - t)e^(-t)j + 6cos(6t)k; at t = 0

= 3(0)²i + (1 - 0)e^(0)j + 6cos(0)k

= j + 6k (Because cos(0) = 1)

r'(0) = j + 6k

r(0) = (2 + t³)i + te^(−t)j + sin(6t)k; at t = 0

= (2 + 0³)i + (0)e^(0)j + sin(0)k

= 2i (Because sin(0) = 0)

r(0) = 2i

Note: Suppose A = xi +yj +zk

|A| = √(x² + y² + z²).

So |r(0)| = √(2²) = 2

And finally, we can obtain the unit tangent vector

r'(0)/|r(0)| = (j + 6k)/2

= j/2 + 3k

8 0
3 years ago
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