Wath is six hundred ninety- two thousand,four

To find the distance AB across a river, a surveyor laid off a distance BC=361 m on one side of the river. It is found that B=111

zhannawk [14.2K]

Answer:

91.87m

Step-by-step explanation:

We are given that

BC=361 m

Angle B= $111^{\circ}20'=111+\frac{20}{60}=111.33^{\circ}$

1 degree= 60 minute

Angle C= $12^{\circ} 15'=12+\frac{15}{60}=12.25^{\circ}$

We have to find the value of AB.

Sum of angles of triangle=180 degrees

$\angle A+\angle B+\angle C=180$

$\angle A=180-(\angle C+\angle B)=180-(111.33+12.25)=56.42^{\circ}$

By using law of sine

$\frac{AB}{sin C}=\frac{BC}{sin A}$

$AB=\frac{BCsin C}{sin A}=\frac{361 sin 12.25^{\circ}}{sin 56.42^{\circ}}$

$AB=91.87 m$

7 0

4 years ago

Im really stressed and i really need help. No one is helping me on my question.... please HELP!!

lorasvet [3.4K]

1) 4 sqrt x^3 =====> 4x^3/2 =====> 4x sqrt x

Steps:

4 sqrt x^3

sqrt x^3

x^3/2

Answer ======> 4x^3/2 =====> 4x sqrt x

2): 1 / x^ -1 =====> x

Steps:

1 / x^ -1

Apply exponent rule:

a^ -1 = 1 / a

x^ - 1 = 1 / x

1 / 1 / x

Apply fraction rule:

1 / b / c = c / b

x / 1

Apply Rule:

a / 1 = a

Answer ========> x

3): 10 sqrt x^5 * x^4 * x^2 = 10^17/2 ====> 100000000 * 10.5 (Decimal: 316227766 ).

Steps:

10 sqrt x^5 * x^4 * x^2

Apply exponent rule:

a^b * a^c = a^b + c

x^4x^2 = x^4 + 2 = x^6

10x^6 sqrt x^5

sqrt x^5 = x^5/2

10x^6x^5/2

Apply exponent rule:

a^b * a^c = a^b + c

x^6x^5/2 = x^5/2 + 6 = x^17/2

Answer ====> 10x^17/2 ==> 100000000 * 10.5 (Decimal: 316227766 ).

4): x^1/3 * x^1/3 * x^1/3 = x^3 / 27 ======> x^1/9

Steps:

x^1/3 * x^1/3 * x^1/3

Apply rule:

a^1 = a

x^1 = x

x/3 * x/3 * x/3

Multiply fractions:

a/b * c/d = a/b * c/d

xxx / 3*3*3

Multiply numbers:

3*3*3 = 27

Answer =========> x^3 / 27 =======> x^1/9

Hope that helps!!!! : )

8 0

4 years ago

Use a system of equations to solve this problem.

Grace [21]

X = amount of seeds.

y = amount of dried fruits.

we know the snack mix contains both, and we know is 10oz, thus x + y = 10, whatever ounces "x" and "y" are.

how much is it for "x" ounces if each one costs $1.5? well is just 1.5*x or 1.5x.

how much is it for "y" ounces if each one costs $2.5? well is just 2.5*x or 2.5x.

the mix contains 10 ounces, each of which costs $2.2 each, how much will it be then for 10 oz? well, is just 10 * 2.2, or $22.

since the whole 10 oz snack mix costs 22 bucks, then 1.5x + 2.5y = 22.

$\bf \begin{cases}
x+y=10\implies \boxed{y}=10-x\\
1.5x+2.5y=22\\
-------------\\
1.5x+2.5\left( \boxed{10-x} \right)=22
\end{cases}
\\\\\\
1.5x+25-2.5x=22\implies -x+25=22\implies 3=x$

how many ounces of dried fruit is there anyway? well, y = 10 - x.

4 0

3 years ago

How do you write 6.7 in two other word forms for a 4th grade assignment

baherus [9]

Six and seven hundredths

7 0

3 years ago

Read 2 more answers

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→1 3x x − 1

Reika [66]

Answer:

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})$ , we are to evaluate it. To evaluate it, we will simply substitute x = 1 into the function since the variable x tends to 1.

$\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= (\dfrac{3(1)}{1-1} - \dfrac{3}{ln1})\\\\= \dfrac{3}{0} - \dfrac{3}{0}\\\\= \infty - \infty (ind)$

Since we got an indeterminate function, we will find the LCM of the function and solve again.

$= \lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= \lim_{x \to 1} \dfrac{3xlnx-3(x-1)}{(x-1)lnx}\\\\\\= \dfrac{3(1)ln(1)-3(1-1)}{(1-1)ln1}\\\\= \frac{3(0)-3(0)}{0(0)} \\\\= \frac{0}{0} (ind)$

Applying L'hospital rule;

$\frac{x}{y} = \lim_{x \to 1} \dfrac{d/dx(3xlnx-3(x-1))}{d/dx((x-1)lnx)}\\\\= \lim_{x \to 1} \dfrac{3x(\frac{1}{x})+ 3lnx-3)}{(x-1)\frac{1}{x} +lnx}\\\\= \lim_{x \to 1} \dfrac{3 + 3lnx-3}{(x-1)\frac{1}{x} +lnx}\\\\= \frac{3ln1}{(1-1)\frac{1}{1} +ln1}\\\\= \frac{0}{0} (ind)$

Applying L'hospital rule again;

$= \lim_{x \to 1} \dfrac{\frac{d}{dx} (3lnx)}{\frac{d}{dx} ((x-1)\frac{1}{x} +lnx)}\\\\= \lim_{x \to 1} \dfrac{\frac{3}{x} }{(x-1)\frac{-1}{x^2} + \frac{1}{x} +\frac{1}{x} }\\\\= \dfrac{\frac{3}{1} }{(1-1)\frac{-1}{1^2} + \frac{1}{1} +\frac{1}{1} }\\\\= \frac{3}{0(-1)+2}\\ \\= \frac{3}{2-0}\\ \\= 3/2$

<em>Hence the limit of the function is 3/2.</em>

7 0

4 years ago