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Natalka [10]
3 years ago
15

Solve this linear equation 2(x+6)<3x+8

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
8 0

Answer:

x > 4

Step-by-step explanation:

Given

2(x + 6) < 3x + 8 ← distribute left side

2x + 12 < 3x + 8 ( subtract 3x from both sides )

- x + 12 < 8 ( subtract 12 from both sides )

- x < - 4

Multiply both sides by - 1, reversing the symbol as a result of multiplying by a negative quantity.

x > 4

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ELEN [110]
The answer is 0.00010224...
5 0
3 years ago
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PLZ ANSWER THIS CORRECTLY FOR 100 POINTS AND BRANLIEST :DD
gavmur [86]

Actually Welcome to the Concept of the Angles,.

Since,the AE is a angle bisector, hence, BAE = EAC simply.

hence we equate them..

===> x+ 30 = 3x - 10

==> 3x - x = 30 + 10

==> 2x = 40

==> x = 20°

hence the major angle EAC = 3(20)-10 = 60-10 = 50°

hence, angle EAC = 50°

8 0
2 years ago
BRAINLIEST!
alisha [4.7K]
Assume that 7 is a. We already know that c = 25. The equation is c^2 - a^2 = b^2.
25^2= 625 7^2= 49 625-49= 576
Square root of 576 = 24

The length of thee other leg is 24.

6 0
3 years ago
Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0
3 years ago
Please help quick i'm confused
Step2247 [10]

Answer:

263.76

Step-by-step explanation:

The formula is pi(3.14) times r(radius squared) (6^2) times h(height)( 7) over 3  

3.14 times 36 times 7/3

3 0
2 years ago
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