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3 years ago

10

Write a rule for the linear function in the graph

1 answer:

bija089 [108]3 years ago

6 0

Ok, so we know the slope-intercept form of a line, y=mx+b, with m being the slope and b being the y-intercept. We know that the y-intercept has to be negative, so that means A is not a viable option. We also know that the slope is up 4 units and to the right 1 unit. Therefore, the slope is (rise over run) +4/+1, or 4. Therefore, C is your best answer.

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If tan θ =(√3)/3, 0◦< θ < 360◦,

svetlana [45]

Arctan (√3 /3) = 30°. = π/6 rad

That is the value searched, in degrees and radians.

You can verifiy that tan(30°) = sin(30°) / cos(30°) = [1/2] / [√3/2] = 1/√3 = √3 / 3

3 0

3 years ago

34.25 multiplied by what Equals 342.5

laiz [17]

Just work backwards.

342.5/ 34.25 = 10

8 0

3 years ago

What is the equation of the line passing through the points (2, 6) and (-2, 4)

g100num [7]

The equation is y=1/2x+5

3 0

3 years ago

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago

If the Federal Reserve buys $100,000 in Treasury bonds from a bank at 3% interest, what is the immediate effect on the money sup

sladkih [1.3K]

$3,000 is the answer

8 0

3 years ago