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saveliy_v [14]
3 years ago
7

A triangle with sides lengths of 5,12 and 13 is a right triangle

Mathematics
2 answers:
djverab [1.8K]3 years ago
7 0
Yep! It’s a right! You got it
antiseptic1488 [7]3 years ago
5 0

Answer:

Yes, it is.

Step-by-step explanation:

Please mark as Brainliest! :)

Have a nice day.

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What are the domain and range for this graph?
RideAnS [48]

ANSWER

c. Domain: [0,18]

Range: [0,360]

EXPLANATION

The domain refers to the values of x for which the function is defined.

The given straight line graph is defined for x=0 to x=18.

Hence the domain is [0,18]

The range refers to the values of y for which x is defined.

The graph is defined for y=0 to y=360.

The range is [0,360]

6 0
3 years ago
Which statement is true? Please help me
amid [387]
The second option is true: 3/6= 4/8
7 0
2 years ago
I need help with number 3
yan [13]
First check whether the point (-6,8) is the solution to any of the equations. To check, just plug in the x and y values of the points into the equation and see if they give you a true statement. 
5(-6)+3(8)=-6
-30+24=-6
-6=-6
That's a true statement so the point is the solution to the first equation.
2(-6)+(8)=-4
-12+8=-4
-4=-4
It is a true statement so the point is a solution for both equations
There are no other solution because lines can only intersect in one or infinite points, but that is only if they are the same lines, which is not true in this circumstance.
A. It is the only solution to the set. 
Hope this helps.
7 0
3 years ago
Read 2 more answers
Climbing ar an indoor rock-climbing gym costs $8 per hour. Lizbeth has a $30 gift card to spend at the gym. Write an algebraic e
Vesnalui [34]

Answer:

30=8x

Step-by-step explanation:

lizabeth cant climb for 4.5 hrs because if you divide 30 by 8 you get 3.75. so Lizabeth can climb for 3.75 hrs using only her $30 gift card

4 0
3 years ago
Find f'(x) for f(x) = cos^2(3x^3).
garik1379 [7]

f(x) = \cos^2(3x^3)\\f'(x) = \frac{d}{dx}[\cos^2(3x^3)]\\= 2\cos(3x^3) \cdot \frac{d}{dx}[cos(3x^3)]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot \frac{d}{dx}[3x^3]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot 9x^2\\= -18x^2\cos(3x^3)\sin(3x^3)

6 0
3 years ago
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