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vladimir2022 [97]
3 years ago
8

The solution to the equation x3 = 125

Mathematics
2 answers:
wariber [46]3 years ago
5 0
You divide 3 to both sides that then will give you a result of
x= 41.66
garri49 [273]3 years ago
3 0

Answer: 5

Step-by-step explanation: Take the cube root of both sides. The cube root of 125 is 5

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Write
IRISSAK [1]

Answer:

y = 2 ( x − 1 ) ^ 2 + 2

Step-by-step explanation:

y = a ( x − h ) ^ 2 + k

where vertex is (h,k) = (1,2)

h=1, k=2

y = a ( x − 1 ) ^ 2 + 2

sub ( 3,10)

10 = a (3 - 1)^2 + 2

a=2

y = 2 ( x − 1 ) ^ 2 + 2

7 0
2 years ago
Help please. solve for x and y.
ohaa [14]
To solve for x and y, you need to know if the triangles are congruent. They are because of their congruent sides and exterior angles are the same measure.
Now you know that the triangles are congruent, you can set up a proportion for the side lengths you are looking for.
For x...

22/10=x/12. Then cross mutiply. x =26.4
You can do this because the triangles are congruent.
For y...

22/10 =37.4/y. Then you cross multiply. y=17

The correct answer is B.
4 0
3 years ago
Please help me please will give brainliest to ​
uranmaximum [27]

Answer:

B.

<h3>step by step explanation</h3>

I hope it's help

7 0
2 years ago
Fill in the table using this function rule
marusya05 [52]
<h3>hello!</h3>

Plug in the values of x and solve for y:-

y=5+4x

y=5+4(4)

y=5+16

y=21

________________________________

y=5+4x

y=5+5(5)

y=5+20

y=25

_________________________________

y=5+4x

y=5+4(8)

y=5+32

y=37

___________________________________

y=5+4x

y=5+4(10)

y=5+40

y=45

__________________________________

<h3>note:-</h3>

Hope everything is clear; if you need any clarification/explanation, kindly let me know, and I will comment and/or edit my answer :)

8 0
2 years ago
Read 2 more answers
-5t^2+10t+22 does the ball reach a height of 35m
Alika [10]

Answer:

Step-by-step explanation:

35=-5t²+10t+22

5t²-10t+13=0

disc =b²-4ac=(-10)²-4×5×13=100-260=-160<0

t is imaginary.

So ball never reaches a height of 35 m

3 0
3 years ago
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