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anastassius [24]
3 years ago
10

Find the missing side of the triangle

Mathematics
1 answer:
Gnom [1K]3 years ago
6 0
Use the Pythagorean theorem.
x^2 + 5^2 = 13^2
x^2 + 25 = 169
x^2 = 144
x = 12
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Translate the following statement into an appropriate algebraic statement: The number of text messages sent times $0.15 plus the
Oxana [17]
Actually the answers A since t is for text messages and it costs .15 and m is minutes and costs .07 so we add those two together to he .15t+ .07m then there's a discount of 2.50 so we subtract. So our final equation is .15t + .07m - 2.50 = C
7 0
3 years ago
Heres the picture, thank you!
Arte-miy333 [17]
Leg * /2 = hypotenuse
Hypotenuse divided by /2 = leg

12 divided by /2

Rationalize denominator:
12 * /2 divided by by /2 * /2

Simplify denominator:
12 * /2 divided by 2

Simplify:
6 * /2 or 4.24

Both x and y because it’s a 45-45-90 triangle
7 0
3 years ago
Read 2 more answers
The function g(x)=(x-2^2. The function f(x)=g(x) +3
OlgaM077 [116]
The parent function is:
 y = x ^ 2
 Applying the following function transformation we have:
 Horizontal translations:
 Suppose that h> 0
 To graph y = f (x-h), move the graph of h units to the right.
 We have then:
 g (x) = (x-2) ^ 2
 Then, we have the following function transformation:
 Vertical translations
 Suppose that k> 0
 To graph y = f (x) + k, move the graph of k units up.
 We have then that the original function is:
 g (x) = (x-2) ^ 2
 Applying the transformation we have
 f (x) = g (x) +3
 f (x) = (x-2) ^ 2 + 3
 Answer:
 
the function f(x)  moves horizontally 2 units rigth.
 
The function f (x) is shifted vertically 3 units up.
4 0
3 years ago
What is the domain of f(x)=9-x²<br> A(.fx)≥9<br> B.All real numbers<br> C.-3≤x≤3<br> D. x≤9
Ber [7]

Answer:

B. all real numbers

Step-by-step explanation:

hope that helps

3 0
3 years ago
Read 2 more answers
HELP ME PLEASE IVE BEEN STRUGGLING WITH THIS FOR SO LONG
Ber [7]

Answer:

<h2>m∠ACE = 90°</h2>

Step-by-step explanation:

Figure Interpretation:

m∠CBA + m∠CDE = 180

m∠BCA = (180-m∠CBA)/2

m∠DCE = (180-m∠CDE)/2

=======================

Then

m∠BCA + m∠DCE = (180-m∠CBA)/2 + (180-m∠CDE)/2

                                = [360-(m∠CBA+m∠CDE)]/2

                                = [360 - 180]/2

                                = 90

finally,

m∠ACE= 180 - (m∠BCA + m∠DCE)

             = 180 - 90

             = 90

8 0
2 years ago
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