Classify the expression by both degree and terms 2x4

Find the value of the length a x rounded to 1 DP. The diagram is not drawn accurately

son4ous [18]

Answer:

x ≈ 3.7 m

Step-by-step explanation:

Use the right triangle on the left to find the height h , using the sine ratio

sin39° = $\frac{opposite}{hypotenuse}$ = $\frac{h}{7}$ ( multiply both sides by 7 )

7 × sin39° = h , then

h = 4.4

Using the tangent ration on the right side right triangle, then

tan50° = $\frac{opposite}{adjacent}$ = $\frac{h}{x}$ = $\frac{4.4}{x}$ ( multiply both sides by x )

x × tan50° = 4.4 ( divide both sides by tan50° )

x = $\frac{4.4}{tan50}$ ≈ 3.7 m ( to 1 dec. place )

8 0

3 years ago

What is the equation of the line in slope-intercept form?

Marysya12 [62]

Answer:

The equation of the line in slope-intercept form is:

y = x + 4

Step-by-step explanation:

The slope-intercept form of the line equation

$y = mx+b$

where

m is the slope

b is the y-intercept

Given the points on the line graph

(0, 4)

(1, 5)

Determining the slope between (0, 4) and (1, 5)

(x₁, y₁) = (0, 4)

(x₂, y₂) = (1, 5)

Using the formula

Slope = m = [y₂ - y₁] / [x₂ - x₁]

= [5 - 4] / [1 - 0]

= 1 / 1

= 1

Thus, the slope of the line = m = 1

We know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

From the graph, it is clear

at x = 0, y = 4

Thus, the y-intercept b = 4

now substituting b = 4 and m = 1 in the slope-intercept form

y = mx + b

y = (1)x + 4

y = x + 4

Therefore, the the equation of the line in slope-intercept form is:

y = x + 4

7 0

3 years ago

Numbers with a quotient of 8

Sever21 [200]

Answers:

1. 48÷6=8

2. 16÷2=8

3. 40÷5= 8

4. 64÷8=8

5. 56÷7= 8

8 0

3 years ago

Read 2 more answers

The line represented by y = 3x − 2 and a line perpendicular to it intersect at R(3, 7). What is the equation of the perpendicula

laiz [17]

Answer:

y= -1/3(x) + 8

Step-by-step explanation:

slope of the 1st line = 3

slope of the line perpendicular to the first is its opposite reciprocal, so

slope of the 2nd line = -1/3

now we have a point and the slope of the perpendicular line, so we use the point slope formula to get:

y-7=-1/3(x-3)

y= -1/3(x) + 1 + 7

y= -1/3(x) + 8

7 0

3 years ago

Consider the system of differential equations dxdt=−4ydydt=−4x. Convert this system to a second order differential equation in y

koban [17]

$\dfrac{\mathrm dy}{\mathrm dt}=-4x\implies x=-\dfrac14\dfrac{\mathrm dy}{\mathrm dt}\implies\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}$

Substituting this into the other ODE gives

$-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}=-4y\implies y''-16y=0$

Since $x(t)=-\dfrac14y'(t)$ , it follows that $x(0)=-\dfrac14y'(0)=4\implies y'(0)=-16$ . The ODE in $y$ has characteristic equation

$r^2-16=0$

with roots $r=\pm4$ , admitting the characteristic solution

$y_c=C_1e^{4t}+C_2e^{-4t}$

From the initial conditions we get

$y(0)=5\implies 5=C_1+C_2$

$y'(0)=16\implies-16=4C_1-4C_2$

$\implies C_1=\dfrac12,C_2=\dfrac92$

So we have

$\boxed{y(t)=\dfrac12e^{4t}+\dfrac92e^{-4t}}$

Take the derivative and multiply it by -1/4 to get the solution for $x(t)$ :

$-\dfrac14y'(t)=\boxed{x(t)=-\dfrac12e^{4t}+\dfrac92e^{-4t}}$

7 0

4 years ago

Classify the expression by both degree and terms 2x4 - x +5