Use triangles ABC and A'B'C' for Exercise. Use the coordinates to find the lengths of the sides.

Please I need help 9x + 14 ≤ 95

Wittaler [7]

ANSWER

My answer is in the photo above

7 0

3 years ago

Read 2 more answers

Evaluate the definite integral. 1 x4(1 + 2x5)5 dx.

Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ , we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

$= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}$

$= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} } \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du } \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}] \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\$

$= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C$

substitute u = 1-2x⁵ into the result

$= \dfrac{1}{40(1-2x^5)^4} +C$

Hence $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ $= \dfrac{1}{40(1-2x^5)^4} +C$

5 0

3 years ago

Prove that 25^11−5^19 is divisible by 31.

SCORPION-xisa [38]

Answer:

Divisible by 3 is the answer

Step-by-step explanation:

First get everything to have the same base of 5

25^11 - 5^19

(5^2)^11 - 5^19

5^(2*11) - 5^19

5^22 - 5^19

Now factor out the GCF 5^19 to get

5^22 - 5^19

5^(19+3) - 5^(19+0)

5^19*5^3 - 5^19*5^0

5^19(5^3 - 5^0)

5^19(125 - 1)

5^19*(124)

At this point, we factor the 124 into 31*4 to end up with this full factorization: 5^19*31*4

Therefore, 25^11 - 5^19 is equivalent to 5^19*31*4

Since 31 is a factor of the original expression, this means the original expression is divisible by 31.

4 0

3 years ago

What is the true solution to 3 in 2 + in 8 = 2 in (4x)

11Alexandr11 [23.1K]

The first step for solving this equation is to determine the defined range.
3㏑(2) + ㏑(8) = 2㏑(4x), x > 0
Write the number 8 in exponential form.
3㏑(2) + ㏑(2³) = 2㏑(4x)
Using ㏑( $a^{x}$ ) = x × ㏑(a),, transform the expression.
3㏑(2) + 3㏑(2) = 2㏑(4x)
Now collect the like terms on the left side of the equation.
6㏑(2) = 2㏑(4x)
Switch the sides of the equation.
2㏑(4x) = 6㏑(2)
Using x × ㏑(a) = ㏑( $a^{x}$ ),, transform the expression on the left side of the equation.
㏑((4x)²) = 6㏑(2)
Using x × ㏑(a) = ㏑( $a^{x}$ ),, transform the expression on the right side of the equation.
㏑((4x)²) = ㏑( $2^{6}$ )
Since the bases of the logarithms are the same,, you need to set the arguments equal.
(4x)² = $2^{6}$
Take the square root of both sides of the equation and remember to use both the positive and negative roots.
4x = +/- 8
Now separate the equation into 2 possible cases.
4x = 8
4x = -8
Solve the top equation for x.
x = 2
Solve the bottom equation for x.
x = 2
, x > 0
x = -2
Lastly,, check if the solution is in the defined range to find your final answer.
x = 2
This means that the correct answer to your question is x = 2.
Let me know if you have any further questions
:)

5 0

3 years ago

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For which values of P and Q does the following equation have infinitely many solutions ?

solmaris [256]

Answer:

(P, Q) = (-75, 57)

Step-by-step explanation:

The equation will have infinitely many solutions when it is a tautology.

Subtract the right side from the equation:

Px +57 -(-75x +Q) = 0

x(P+75) +(57 -Q) = 0

This will be a tautology (0=0) when ...

P+75 = 0

P = -75

and

57-Q = 0

57 = Q

_____

These values in the original equation make it ...

-75x +57 = -75x +57 . . . . . a tautology, always true

4 0

3 years ago