Find the value of x. Then classify the triangle

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 x. 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,

$f(x)=\ln x+ \ln 5^2 +\ln1\\
f(x)=\ln x+ \ln 25 +\ln1$

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

$f(x)=0+\ln 25 +\ln1\\
f(x)=0+\ln 25 +\ln1\\
f(x)=\ln 25 +\ln1\\
f(x)=\ln 25 +0\\
f(x)=\ln 25$

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:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

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

Algebra I

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

Step-by-step explanation:

Step 1: Define Systems

y = x - 16

5y = 2x - 56

Step 2: Solve for x

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

Step 3: Solve for y

Define original equation: y = x - 16
Substitute in x: y = 8 - 16
Subtract: y = -8

5 0

3 years ago