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3 years ago

If you're reproducing a 4 cm x 4 cm picture on a 3 mx 3 m wall, what ratio should you use?

Mathematics

1 answer:

Svetlanka [38]3 years ago

4 0

Answer:

hi stranger

Step-by-step explanation:

do u need some help

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Math problem, please help:)

alukav5142 [94]

3_/—- 5 i’m pretty sure

5 0

3 years ago

Please help. Thank you

shepuryov [24]

complementary is 90 degrees. so, you combine like terms.

7 0

3 years ago

Complete the table of inputs and outputs for the given function g(x) = 3 - 8x

NeX [460]

Answer:

See Explanation

Step-by-step explanation:

<em>The question is incomplete as the input and output values are not given. However, the question can still be solved by assuming both values</em>

Given

$g(x) = 3 - 8x$

In this case;

x represents the input while g(x) represents the output

Now, assume x = 2; Determine the output.

Here we simply substitute 2 for x in the above expression

i.e.

$g(x) = 3 - 8x$ becomes

$g(2) = 3 - 8(2)$

$g(2) = 3 - 16$

$g(2) = -13$

Similarly, if x = 5

$g(5) = 3 - 8(5)$

$g(5) = 3 - 40$

$g(5) = -37$

Also, let's assume g(x) = -21; Determine the input

$g(x) = 3 - 8x$ becomes

$-21 = 3 - 8x$

Solve for x

$-21 - 3 = -8x$

$-24 = -8x$

$x = \frac{-24}{-8}$

$x = 3$

7 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

A rectangle has a height of 7 and a width of 2x2 – 3.

yanalaym [24]

Answer:14^2-21

Step-by-step explanation:

5 0

2 years ago