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tangare [24]
2 years ago
12

The length of a square is 3x - 4 units. Find the area in terms of x.

Mathematics
2 answers:
Alja [10]2 years ago
8 0

Answer:

9x^2 - 24x + 16 is area in terms of x alternativly (3x-4)^2 is also true

Step-by-step explanation:

L * W =  area  length = width on a sqare so L * L = area too

(3x - 4) * (3x - 4)

use foil method

9x^2 - 24x + 16

storchak [24]2 years ago
6 0

Answer:

9x^2 − 24x + 16

Step-by-step explanation:

To find the area of a square, you use the formula

A = x^2

So we plug in 3x-4 for x

A = (3x-4)^2

A = 9x^2 − 24x + 16

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Answer:

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Step-by-step explanation:

6 0
3 years ago
What is the correct method to label a segment with endpoint E and endpoint F? 
Degger [83]

Answer : option B

the correct method to label a segment with endpoint E and endpoint F

The endpoints of the line are E  and F

The Line starts at  E  and ends at F. The line will not extend after F

The line having end points are called as line segment

We represent a line segment by putting a line at the top

to label a segment with endpoint E and endpoint F, we use EF  or FE and a line at the top

So answer is option B

4 0
3 years ago
Read 2 more answers
If 2tanA=3tanB then prove that,<br>tan(A+B)= 5sin2B/5cos2B-1​
Fed [463]

By definition of tangent,

tan(A + B) = sin(A + B) / cos(A + B)

Using the angle sum identities for sine and cosine,

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

yields

tan(A + B) = (sin(A) cos(B) + cos(A) sin(B)) / (cos(A) cos(B) - sin(A) sin(B))

Multiplying the right side by 1/(cos(A) cos(B)) uniformly gives

tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) tan(B))

Since 2 tan(A) = 3 tan(B), it follows that

tan(A + B) = (3/2 tan(B) + tan(B)) / (1 - 3/2 tan²(B))

… = 5 tan(B) / (2 - 3 tan²(B))

Putting everything back in terms of sin and cos gives

tan(A + B) = (5 sin(B)/cos(B)) / (2 - 3 sin²(B)/cos²(B))

Multiplying uniformly by cos²(B) gives

tan(A + B) = 5 sin(B) cos(B) / (2 cos²(B) - 3 sin²(B))

Recall the double angle identities for sin and cos:

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

cos(2x) = cos²(x) - sin²(x)

and multiplying uniformly by 2, we find that

tan(A + B) = 10 sin(B) cos(B) / (4 cos²(B) - 6 sin²(B))

… = 10 sin(B) cos(B) / (4 (cos²(B) - sin²(B)) - 2 sin²(B))

… = 5 sin(2B) / (4 cos(2B) - 2 sin²(B))

The Pythagorean identity,

cos²(x) + sin²(x) = 1

lets us rewrite the double angle identity for cos as

cos(2x) = 1 - 2 sin²(x)

so it follows that

tan(A + B) = 5 sin(2B) / (4 cos(2B) + 1 - 2 sin²(B) - 1)

… = 5 sin(2B) / (4 cos(2B) + cos(2B) - 1)

… = 5 sin(2B) / (4 cos(2B) - 1)

as required.

5 0
2 years ago
What is the product of x+2 and x-7? Write your answer in standard form.
Leto [7]
X^2-5x-14 
It is the correct answer I think
5 0
3 years ago
I need to give an equation for an absolute value function that has a minimum. How do I determine that it has a minimum?.
tatyana61 [14]
To determine the minimum of an equation, we derive the <span>equation using differential calculus twice (or simply </span><span>take the second derivative of the function). If the </span><span>second derivative is greater than 0, then it is minimum; </span><span>else, if it is less than 1, the function contains the </span><span>maximum. If the second derivative is zero, then the </span><span>inflection point </span><span>is</span><span> identified.</span>
5 0
3 years ago
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