A mailbox casts a 4-foot shadow. A nearby tree that measures 12 feet casts an 15-foot shadow. How tall is the mailbox?

<img src="https://tex.z-dn.net/?f=15%20x%5E%7B2%7D%20%2B14x-49%3D0" id="TexFormula1" title="15 x^{2} +14x-49=0" alt="15 x^{2} +1

Alex777 [14]

$15x^2+14x-49=0\\\\
Terms:\\
a=15\ \ \ \ b=14\ \ \ \ c=-49\\
Quadratic\ Formula:\\\\
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\
x=\dfrac{-14\pm\sqrt{196+2,940}}{30}\\\\
x=\dfrac{-14\pm\sqrt{3,136}}{30}\\\\
x=\dfrac{-14\pm56}{30}$

$Roots:\\\\
x'=\dfrac{-14+56}{30}\\\\
x'=\dfrac{42}{30}\\\\
x'=\dfrac{21}{15}\\\\
\boxed{x'=\dfrac{7}{5}}\\\\\\\\
x''=\dfrac{-14-56}{30}\\\\
x''=\dfrac{-70}{30}\\\\
\boxed{x''=\dfrac{-7}{3}}$

8 0

3 years ago

Read 2 more answers

Based on Pythagorean identities, which equation is true? A. Sin^2 theta -1= cos^2 theta B. Sec^2 theta-tan^2 theta= -1 C. -cos^2

Arturiano [62]

Answer:

D

Step-by-step explanation:

our basic Pythagorean identity is cos²(x) + sin²(x) = 1

we can derive the 2 other using the listed above.

1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

1 + tan²(x) = sec²(x)

2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)

cot²(x) + 1 = csc²(x)

A. sin^2 theta -1= cos^2 theta

this is false

cos²(x) + sin²(x) = 1

isolating cos²(x)

cos²(x) = 1-sin²(x), not equal to sin²(x)-1

B. Sec^2 theta-tan^2 theta= -1

1 + tan²(x) = sec²(x)

sec²(x)-tan(x) = 1, not -1

false

C. -cos^2 theta-1= sin^2

cos²(x) + sin²(x) = 1

sin²(x) = 1-cos²(x), our 1 is positive not negative, so false

D. Cot^2 theta - csc^2 theta=-1

cot²(x) + 1 = csc²(x)

isolating 1

1 = csc²(x) - cot²(x)

multiplying both sides by -1

-1 = cot²(x) - csc²(x)

TRUE

3 0

3 years ago

Find the value of 2a squared + 5b squared when a = -6 and b =2

jenyasd209 [6]

92

given 2a² + 5b²

substitute given values for a and b into the expression

2(- 6)² + 5(2)² = (2 × 36 ) + (5 × 4) = 72 + 20 = 92

5 0

3 years ago

I need help 6x-2y=10 x-2y=-5 solve by elimination

dem82 [27]

<h3><u>Explanation</u></h3>

Given the system of equations.

$\begin{cases} 6x - 2y = 10 \\ x - 2y = - 5 \end{cases}$

Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

$\begin{cases} 6x - 2y = 10 \\ - x + 2y = 5 \end{cases}$

There as we can get rid of the y-term by adding both equations.

$(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x = \frac{15}{5} \longrightarrow \frac{ \cancel{15}}{ \cancel{5}} = \frac{3}{1} \\ x = 3$

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

$x - 2y = - 5 \\ 3 - 2y = - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\ \frac{8}{2} = y \\ y = \frac{8}{2} \longrightarrow \frac{ \cancel{8}}{ \cancel{2}} = \frac{4}{1} \\ y = 4$

Hence, the value of y is 4. Therefore, we can say that when x = 3, y = 4.

Answer Check by substituting both x and y values in both equations.

<u>First</u><u> </u><u>Equation</u>

$6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10 = 10 \longrightarrow \sf{true} \: \green{ \checkmark}$

<u>Second</u><u> </u><u>Equation</u>

$x - 2y = - 5 \\ 3 - 2(4) = - 5 \\ 3 - 8 = - 5 \\ - 5 = - 5 \longrightarrow \sf{true} \: \green{ \checkmark}$

Hence, both equations are true for x = 3 and y = 4. Therefore, the solution is (3,4)

<h3><u>Answer</u></h3>

$\begin{cases} x = 3 \\ y = 4 \end{cases} \\ \sf \underline{Coordinate \: \: Form} \\ (3,4)$

8 0

3 years ago

Add the following military times: 1355 + 0540

Mnenie [13.5K]

If its multiplcation
731700

if its addition
1895

im sorry if im wrong but im just trying to help out and please correct me if ime wrong

4 0

3 years ago