Does this graph show a function? Explain how you know.

A man walks 4km from point x due east of point Y. The bearing of a flag point x and y are due N80°W and N40°E respectively. Find

In-s [12.5K]

The distance of the flag pole from point Y is 0.8 km and it is placed at an bearing of N40°E

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Sine rule shows the relationship between the sides and angles of a triangle.

The triangle formed has angles A = 10°, B = 50°, C = 120°, c = 4 km, a = distance from point y.

Hence:

a / sinA = c / sinC

a / sin(10) = 4 / sin(120)

a = 0.8 km

The distance of the flag pole from point Y is 0.8 km and it is placed at an bearing of N40°E

Find out more on equation at: brainly.com/question/2972832

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3 0

2 years ago

Ryan has 76,406. Reggie has 55,998. About how many more marbles does Ryan have than Reggie?

Murrr4er [49]

the difference is 20,408

7 0

3 years ago

Read 2 more answers

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago

Helppppppppppppppppp

sweet-ann [11.9K]

Answer:

the answer is 112

Step-by-step explanation:

7² = 49.

4² = 16.

49+3(16+3+2) = 49+3(21) = 49+63 = 112

3 0

3 years ago

Read 2 more answers

Solve for a and b. please help.

Pachacha [2.7K]

Answer:

a= 22.5

b= 37.5

Step-by-step explanation:

In △BCD:

Applying Pythagoras' Theorem,

a² +30²= b²

a² +900= b² -----(1)

In △ABC:

Applying Pythagoras' Theorem,

b² +50²= (40 +a)² -----(2)

Substitute (1) into (2):

a² +900 +50²= 40² +2(40)(a) +a² (expand bracket)

a² +900 +2500= 1600 +80a +a²

a² +3400= a² +80a +1600 (simplify)

a² +3400 -a² -80a -1600= 0 (bring everything to 1 side)

-80a +1800= 0

80a= 1800 (+80 on both sides)

a= 1800 ÷80

a= 22.5

Subst. a= 22.5 into (1):

22.5² +900= b²

b²= 506.25 +900

b²= 1406.25

b= √1406.25 (square root both sides)

b= 37.5 (reject negative value since b is a length)

☆(a +b)²= a² +2ab +b²

3 0

3 years ago