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Naya [18.7K]
3 years ago
7

HELLLLPPPPPPPP PLEASEEEEEEEEE

Mathematics
1 answer:
tatyana61 [14]3 years ago
4 0

Answer:

10.5

Step-by-step explanation:

KET\cong KIT...... (given)

\therefore KE \cong KI.... (csct)

\therefore m(KE) =m(KI)

\therefore 10x-34= 4x + 29

\therefore 10x-4x= 34 + 29

\therefore 6x= 63

\therefore x=\frac{63}{6}

\therefore x=10.5

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X2=9 has only one solution, which is 4.5
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Solutions to -8x + 5 less than or equal to 11
Vilka [71]

Answer:

x\geq -\frac{3}{4}

Step-by-step explanation:

-8x+5\leq 11\\\\1. -8x\leq 11-5\\   -8x\leq 6

Isolate the parts of the equation with the x variable

2. \frac{-8}{-8} x\geq\frac{6}{-8}  \\x\geq -\frac{3}{4}

When you divide or multiply by a negative number in inequalities, you must flip the sign of the inequality.

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Cheese sticks that were previously priced at "10 for $1" are now "4 for $1". Find each percent change.
statuscvo [17]

original quantity : 10 for $1

new quantity: 4 for $1

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10-4=6

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0.06=6%

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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
PLZ ANSWER ASAP PLZ Which is a solution to the equation 4b2 − 11b = 28 + 5b2? A) −5 B) −4 C) −3 D) 4
Dovator [93]
D) 4
i think this is right
7 0
3 years ago
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