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ANTONII [103]
3 years ago
10

Won't you answer a poor sinner's question HELP! I SWEAR I'LL MARK IT AS BRAINIEST OR SMTHN​

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

you needed to mark this question for the science section

Step-by-step explanation:

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Marissa rode her bike to the oark and back home. She lives 2 3/10 miles from the park . How many miles did Marissa ride her bike
Oduvanchick [21]

will she would have rode 4 3/5 miles in all.

2 3/10+2 3/10=4 3/5

that is 2 3/10 there and 2 3/10 back so you would add that together.

MILES IN ALL 4 3/5

Hope this helps

3 0
3 years ago
The rule that best supports your answer to Question 5 would be:
mixas84 [53]
Sense the two triangles are congruent you would use CPCTC.
6 0
4 years ago
The sides of a right triangle are x, x+21, and x+27. what is the length of the hypotenuse
mina [271]
Answer: 51 uinits

The three numbers are x , x + 21 and x + 27. 
The greatest of the 3 numbers is x + 27, so x + 27 has to be the hypotenuse.

----------------------------------------------
Pythagoras Theorem
----------------------------------------------

a² + b²  = c²

x² + (x+21)² = (x+27)²

x² + x² + 42x + 441 = x² + 54x + 729

x² - 12x -288 = 0

(x - 24)(x + 12) = 0

x = 24 or x = -12 (rejected, length cannot be negative)

----------------------------------------------
Find hypotenuse
----------------------------------------------
 Hypotenuse = x + 27 = 24 + 27 = 51 units

-----------------------------------------------
Answer: 51 units
----------------------------------------------

3 0
4 years ago
Read 2 more answers
Use the quadratic formula to solve x2 + 2x - 120 = 0. Think about what other method could also be used to solve this equation. a
pshichka [43]

Answer:

x = 10 , -12

Step-by-step explanation:

Solution:-

- The given quadratic equation is to be solved using the quadratic formula. The general form of a quadratic equation is:

                           ax^2 + bx + c =0  

Where, [ a , b and c are constants ]

- The quadratic formula is given as:

                       

                           x = \frac{-b+/-\sqrt{b^2 - 4*a*c} }{2a}

- The given equation is:

                             x^2 + 2x - 120 = 0

Where, a = 1 , b = 2 , c = -120

- Solve using quadratic formula:

                       x = \frac{-2+/-\sqrt{2^2 - 4*1*(-120)} }{2*1}\\\\x = \frac{-2+/-\sqrt{4 + 480} }{2}\\\\x = \frac{-2+/-(22) }{2}\\\\\\x = 10 , -12    

7 0
3 years ago
Consider these four statements about a line that passes through two points on a plane.
Brilliant_brown [7]
I think is statement B
8 0
4 years ago
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