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soldi70 [24.7K]
3 years ago
15

Find the value of x. Then classify the triangle

Mathematics
1 answer:
vredina [299]3 years ago
5 0
So basically what you have to do is
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H(0)=
Goshia [24]

Answer:

I feel like it b hope it correct though

6 0
3 years ago
This year Heather watched 20 movies. She thought that 9 of them were very good. Of the movies she watched, what percentage did s
Alex17521 [72]

Answer:

45%

Step-by-step explanation:

PLEASE MARK ME AS BRAINLIEST I REALLY WANT TO LEVEL UP

6 0
2 years ago
Which expressions are equivalent to ln x 2 ln 5 ln 1? Check all that apply. 2 ln 5x ln (x 26) ln 25x ln 1 ln 25x.
Marysya12 [62]

Equivalent expression is the expression of two equation when the way of representation of the two equation is different but the result is same.The expressions which are equivalent to the given expression are,

  • \ln25+\ln1
  • \ln25

Given information-

The expression given in the problem is,

\ln x+2 \ln 5 +\ln1

<h3>Equivalent expression-</h3>

Equivalent expression is the expression of two equation when the way of representation of the two equation is different but the result is same.

For given expression,

\ln x+2 \ln 5 +\ln1

As the above function is the function of <em>x. </em>Thus it can be written as,

f(x)=\ln x+2 \ln 5 +\ln1

<h3>Logarithm power rule</h3>

Logarithm power rule states that the exponent of the logarithm function can transfers to front of the logarithm and vice versa. Thus,

f(x)=\ln x+ \ln 5^2 +\ln1\\&#10;f(x)=\ln x+ \ln 25 +\ln1

Now the value of \ln 1 is equal to the zero. Thus,

f(x)=0+\ln 25 +\ln1\\&#10;f(x)=0+\ln 25 +\ln1\\&#10;f(x)=\ln 25 +\ln1\\&#10;f(x)=\ln 25 +0\\&#10;f(x)=\ln 25

Hence the expressions which are equivalent to the given expression are,

  • \ln25+\ln1
  • \ln25

Learn more about the equivalent expression here;

brainly.com/question/10628562

5 0
2 years ago
Read 2 more answers
Find an explicit solution of the given initial-value problem. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0
MissTica

Answer:

a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x*(1 + 4y²) dx = 0

(1 + x⁴) dy  = - x*(1 + 4y²) dx

[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx

∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx

now to solve each integral

I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁

I₂=  ∫[-x/(1 + x⁴)] dx

for u= x² → du=x*dx

I₂=  ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ =  - tan⁻¹ (x²) +C₂

then

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C

for y(x=1) = 0

1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C

since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for  π*N , we will choose for simplicity N=0 . hen an explicit solution would be

1/2 * 0 = - π/4 + C

C= π/4

therefore

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

5 0
3 years ago
PLEASE HELP MEE!! SHOW YOUR WORK PLS
sleet_krkn [62]

Answer:

(8, -8)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

  • Terms/Coefficients
  • Coordinates (x, y)
  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

y = x - 16

5y = 2x - 56

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute in <em>y</em>:                                                                                                5(x - 16) = 2x - 56
  2. Distribute 5:                                                                                                     5x - 80 = 2x - 56
  3. [Subtraction Property of Equality] Subtract 2x on both sides:                     3x - 80 = -56
  4. [Addition Property of Equality] Add 80 on both sides:                                3x = 24
  5. [Division Property of Equality] Divide 3 on both sides:                                x = 8

<u>Step 3: Solve for </u><em><u>y</u></em>

  1. Define original equation:                                                                               y = x - 16
  2. Substitute in <em>x</em>:                                                                                                y = 8 - 16
  3. Subtract:                                                                                                          y = -8
5 0
3 years ago
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