5x + 2 y = 7 4x + y = 8

Anna is making a banner out of 4 congruent triangles as shown below. How much blue trim will she need for each side

Ede4ka [16]

Answer:

The length of each blue trim is 17.2 inches

Step-by-step explanation:

Given

See attachment for banner

Required

The length of each blue trim

Since all 4 triangles are equal, then the dimension of 1 triangle is:

$Side\ 1 = 10$

$Side\ 2 = 14$

The hypotenuse (x) of the triangle blue trim is represented by the blue trim

So:

$x^2 = Side\ 1^2 + Side\ 2^2$

This gives

$x^2 = 10^2 + 14^2$

$x^2 = 100 + 196$

$x^2 = 296$

Take square root

$x = \sqrt{296$

$x = 17.2in$

3 0

3 years ago

Someone please help me

allsm [11]

Answer:

Solutions: $x = \frac{-3}{ 4} + i \sqrt{39}$ , $x = \frac{-3}{4} - i \sqrt{39}$

Step-by-step explanation:

Given the quadratic equation, 2x² + 3x + 6 = 0, where a =2, b = 3, and c = 6:

Use the quadratic equation and substitute the values for a, b, and c to solve for the solutions:

$x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}$

$x = \frac{-3 +/- \sqrt{3^{2} - 4(2)(6)} }{2(2)}$

$x = \frac{-3 +/- \sqrt{9- 48} }{4}$

$x = \frac{-3 +/- \sqrt{-39} }{4}$

$x = \frac{-3 + i \sqrt{39} }{4}$ , $x = \frac{-3 - i \sqrt{39} }{4}$

Therefore, the solutions to the given quadratic equation are:

$x = -\frac{3}{ 4} + i \sqrt{39}$ , $x = -\frac{3}{4} - i \sqrt{39}$

4 0

2 years ago

Arithmetic and Geometric Sequences (Context)

V125BC [204]

The formula for compound interest

A = P( 1 + r/n) ^ (nt)

A is the amount in the account at the end

P is the principal balance or the amount initially invested

r is the annual interest rate in decimal form

n is the number of times it is coupounded per year

t is the number of years

A = 1800 ( 1+ .0375/1) ^ (1*6)

A = 1800 ( 1.0375)^6

A = 2244.92138

Rounding to the nearest cent

A = 2244.92

7 0

1 year ago

Y=arccos(1/x) Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago

What is the value of b in the equation (y^-9)^b=y^45?

nevsk [136]

Answer:

b = (log(y^45))/log(1/y^9) + (2 i π n)/log(1/y^9) for n element Z

Step-by-step explanation:

Solve for b:

(1/y^9)^b = y^45

Take the logarithm base 1/y^9 of both sides:

Answer: b = (log(y^45))/log(1/y^9) + (2 i π n)/log(1/y^9) for n element Z

5 0

4 years ago