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Svetllana [295]
3 years ago
14

Find the exact value of x. Do the side lengths form a Pythagorean triple?

Mathematics
2 answers:
grandymaker [24]3 years ago
7 0

Answer:

Step-by-step explanation:

4^2 + 6^2 = x^2

sqrt 52 = x

x= 7.211

No, it is not a pythagorean triple because x is not a whole number.

Lilit [14]3 years ago
4 0

Answer: Square root (52), No

Step-by-step explanation:

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2x^5 + 4x^4 – x^3 - x^2 + 7 is divided by 2x^2 - 1.
mixer [17]

Answer: 11

Step-by-step explanation: 2x^5 + 4x^4 – x^3 - x^2 + 7= 11

2x^2 - 1= 1

2x^5 + 4x^4 – x^3 - x^2 + 7 is divided by 2x^2 - 1.=11

4 0
3 years ago
HELLO!! :) CAN U PLEASE HELP ME WITH THIS PROBLEM? I WILL GIVE YOU BRAINLIEST AND A THANKS!! PLEASE EXPLAIN TOO!!
miskamm [114]

Answer:

21square feet

Step-by-step explanation:

Count the whole squares first.

if each square is one square foot then 18 whole squares will be 18×1= 18 square feet

add halves (aproximate halves) together. (3 in total)

Therefore A= 18 +3

= 21 square feet

3 0
2 years ago
Read 2 more answers
Solution the the system of equations 2x+4y=-8 and -2x+3y=29
Zina [86]
The answer to this is y=3 and x=-10.
5 0
3 years ago
PLease help me i will give brainliest
Feliz [49]
B is correct the devotion is only 2 compared to 4.

A is wrong because variate A has higher trees then variate B
C is wrong because deviation for A is 4 compared to 2 ( opposite of B)
D is wrong because 4th one is 10 for A and 13 for B. 10 is not greater than 13
7 0
2 years ago
Please help me I need help please​
MakcuM [25]

Answer:

p = 4

Step-by-step explanation:

Given equation:

x^2+(p-3)y^2-4x+6y-16=0

<u>Standard equation of a circle:</u>

(x-a)^2+(y-b)^2=r^2

(where (a,b) is the centre of the circle, and r is the radius)

If you expand this equation, you will see that the coefficient of y^2 is always one.

Therefore, p-3=1

\implies p=1+3=4

<u>Additional information</u>

To rewrite the given equation in the standard form:

\implies x^2+y^2-4x+6y-16=0

\implies x^2-4x+y^2+6y=16

\implies (x-2)^2-4+(y+3)^2-9=16

\implies (x-2)^2+(y+3)^2=16+4+9

\implies (x-2)^2+(y+3)^2=29

So this is a circle with centre (2, -3) and radius √29

3 0
2 years ago
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