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2 years ago

5

The area of a square lawn is 196 yd2. What is the perimeter of the lawn?

1 answer:

Lesechka [4]2 years ago

8 0

Answer:

The correct answer is 14 yards.

Step-by-step explanation:

Area can be described as a measuring unit to measure the space in any flat surface. The square is a flat surface comprising of four equal sides. Area of a square would be the measure of the length of each size.

As per the question, the area of lawn is 196 yards and it is square shaped. This means that the length of each side would be would be 14 yards.

The area will be calculated by taking the square root of one of the equal sides. The square root of 14 is 196. hence, the correct answer is 14 yards.

- Sophia

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Estimate the solution to the system of equations.

Leona [35]

Answer:

$\sf C) \ x = -1\dfrac{2}{5} , \ y = \ 1\dfrac{3}{5}$

Given equations:

a) -3x + 3y = 9

b) 2x - 7y = -14

<u>Make x the subject in equation 1</u>:

-3x + 3y = 9

-3x = 9 - 3y

x = y - 3

<u>Substitute this into equation 2</u>:

2(y - 3) - 7y = -14

2y - 7y = -14 + 6

-5y = -8

y = -8/-5

y = 1.6

Then x = y - 3

= 1.6 - 3

= -1.4

Solution: (x, y) ⇒ (-1.4, 1.6)

3 0

1 year ago

Solve 7.8e^((x/3)ln(5))=14. What are the exact and approximate solutions?

Dimas [21]

$\bold{\text{Answer}:\quad x=3\bigg(\dfrac{ln\frac{14}{7.8}}{ln(5)}\bigg)=1.09}$

<u>Step-by-step explanation:</u>

$7.8e^{\frac{x}{3}\cdot ln(5)}=14\\\\\\e^{\frac{x}{3}\cdot ln(5)}=\dfrac{14}{7.8}\\\\\\ln\bigg(e^{\frac{x}{3}\cdot ln(5)}\bigg)=ln\bigg(\dfrac{14}{7.8}\bigg)\\\\\\\frac{x}{3}}\cdot ln(5)=ln\bigg(\dfrac{14}{7.8}\bigg)\\\\\\\large\boxed{x=3\bigg(\dfrac{ln\frac{14}{7.8}}{ln(5)}\bigg)}\\$

x ≈ 1.09

6 0

2 years ago

Show that √(1-cos A/1+cos A) =cosec A - cot A

Gennadij [26K]

Hi there!

$\sqrt{\frac{1-cosA}{1+cosA}} =$

We can begin by multiplying by its conjugate:

$\sqrt{\frac{1-cosA}{1+cosA}} * \sqrt{\frac{1+cosA}{1+cosA}} = \\\\\sqrt{\frac{(1-cosA)(1 + cosA)}{(1+cosA)(1 + cosA)}} =$

Simplify using the identity:

$1 - cos^2A = sin^2A$

$\sqrt{\frac{(1-cos^2A)}{(1+cosA)^2}} =\\\\\sqrt{\frac{(sin^2A)}{(1+cosA)^2}} =$

Take the square root of the expression:

${\frac{sinA}{1+cosA} =$

Multiply again by the conjugate to get a SINGLE term in the denominator:

${\frac{sinA}{1+cosA} * {\frac{1-cosA}{1-cosA} =\\$

Simplify:

${\frac{sinA(1-cosA)}{1-cos^2A} =$

Use the above trig identity one more:

${\frac{sinA(1-cosA)}{sin^2A} =$

Cancel out sinA:

${\frac{(1-cosA)}{sinA} =$

Split the fraction into two:

${\frac{1}{sinA} - \frac{cosA}{sinA} =$

Recall:

$1/sinA = cscA\\\\cosA/sinA = cotA$

Simplify:

$\frac{1}{sinA} + \frac{cosA}{sinA} = \boxed{cscA - cotA}$

5 0

1 year ago

Let f(x) = x^2 + 6x.

rusak2 [61]

Answer:

(A) The slope of secant line is 18.

(B) The slope of secant line is h+16.

Step-by-step explanation:

(A)

The given function is

$f(x)=x^2+6x$

At x=3,

$f(3)=(3)^2+6(3)=27$

At x=9,

$f(9)=(9)^2+6(9)=135$

The secant line joining (3,27) and (9,135). So, the slope of secant line is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\dfrac{135-27}{9-3}=18$

The slope of secant line is 18.

(B)

The given function is

$f(x)=x^2+6x$

At x=5,

$f(5)=(5)^2+6(5)=55$

At x=5+h,

$f(5+h)=(5+h)^2+6(5+h)=h^2 + 16 h + 55$

The secant line joining (5,55) and $(5+h,h^2 + 16 h + 55)$ . So, the slope of secant line is

$m=\dfrac{h^2 + 16 h + 55-55}{5+h-5}$

$m=\dfrac{h^2 + 16 h }{h}$

$m=h+16$

The slope of secant line is h+16.

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%

AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

7 0

2 years ago