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4 years ago

State the independent variable and dependent variable and find the slope.

1 answer:

4 0

Time is your independent variable, and snowfall is your dependent variable.

Slope can be found by calculating ∆y/∆x:
$x = \frac{0.06 - 0.02}{3 - 1} \\ x = \frac{0.04}{2} \\ x = 0.02$

Snowfall will accumulate 0.02 m every hour.

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In the diagram above,PQR=125°, QRS=r,RST=80°and STU=44°. Calculate the value of r

Elina [12.6K]

Answer:

∆ PQR is congruent to ∆ STU

This means --

P = S

Q = T

R = U

R = 80°

S = P = 70°

So in ∆PQR, we know R = 80° , P = 70° and Q = ?

Using angle sum property of triangle,

P+ Q + R = 180°

70° + 80° + Q = 180°

150° + Q = 180°

Q = 30°

Hope This Helps :D

6 0

3 years ago

The graph of g(x) is the graph of f(x)=5x+15 compressed horizontally by a factor of 1/5 .

Trava [24]

We can easily get the function of g when looking at the graph, but sadly you don't attached any graph in this question.
f(x) = 5x + 15
y = 5x + 15 (f(x) is also equal to y)
0 = 5x + 15 (change the y into 0)
5x = 15
x = 15/3
x = 3 or 0 = x + 3

so basically g(x) = x +3

The answer is C. g(x) x + 3

6 0

3 years ago

Find the value of x.

Bad White [126]

Answer:

Step-by-step explanation:

Bonsoir,

63°

4 0

3 years ago

Read 2 more answers

How many 2/5 teaspoon doses of medicine are in a container that holds 18/10 teaspoon of medicine?

Volgvan

Answer: 4.5

Step-by-step explanation:

3 0

3 years ago

Find the work done by the force field f(x, y, z) = x − y2, y − z2, z − x2 on a particle that moves along the line segment from (

Veronika [31]

The path can be parameterized by $\mathbf r(t)=\langle0,0,1\rangle(1-t)+\langle3,1,0\rangle t=\langle3t,t,1-t\rangle$ with $0\le t\le1$ . So, the work done by $\mathbf F(x,y,z)=\langle x-y^2,y-z^2,z-x^2\rangle$ is given by the line integral

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r$

where $C$ is the line segment. Equivalently,

$\displaystyle\int_{t=0}^{t=1}\langle3t-t^2,t-(1-t)^2,1-t-9t^2\rangle\cdot\langle3,1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(5t^2+13t-2)\,\mathrm dt$
$=\dfrac{37}6$

3 0

3 years ago