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3 years ago

What is the equation of the line shown in this graph? 29 points!

Mathematics

1 answer:

Rudiy273 years ago

8 0

Answer:

Could you give the question in details please?

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Write the equation of a circle with a center (-3,9) and a radius of 14

babymother [125]

Answer:

(x+3) + (y-9) = 14^2

4 0

3 years ago

Find an equation for the line below.<br><br> The points are 5,-2 and -5,-6

horsena [70]

Answer:

y=1/2x-4.5

Step-by-step explanation:

y=mx+b

4 0

3 years ago

The table below shows data for a class's mid-term and final exams:

lina2011 [118]

Mid term :
Q1 = (88 + 85)/2 = 86.5
Q2 = (92 + 95)/2 = 93.5
Q3 = 100
IQR = Q3 - Q1 = 100 - 86.5 = 13.5

final exams :
Q1 = (65 + 78)/2 = 71.5
Q2 = (88 + 82)/2 = 85
Q3 = (95 + 93)/2 = 94
IQR = Q3 - Q1 = 94 - 71.5 = 22.5

so the final exams has the largest IQR

5 0

3 years ago

Read 2 more answers

You are working for a company that creates tubes that measure 10 cm long and have a diameter of 6 cm. The tubes need to be cut j

Nataly_w [17]

Answer:

$A= 2\pi (3cm)^2 +1 2\pi (3cm)(10 cm)= 18 \pi + 60\pi = 78 \pi cm^2$

And replacing the value of $\pi =3.14$ we got:

$A = 78*3.14 cm^2= 244.92 cm^2$

Step-by-step explanation:

For this case we have the long who represent the height 10 cm and the diameter is 6cm. So then the radius is given by:

$r = \frac{D}{2}= \frac{6cm}{2}= 3cm$

And the surface area for a cylinder is given by this formula:

$A= 2\pi r^2 +2 \pi rh$

And replacing the info given we got:

$A= 2\pi (3cm)^2 +1 2\pi (3cm)(10 cm)= 18 \pi + 60\pi = 78 \pi cm^2$

And replacing the value of $\pi =3.14$ we got:

$A = 78*3.14 cm^2= 244.92 cm^2$

4 0

3 years ago

If cos(θ)=2853 with θin Q IV, what is sin(θ)?

marysya [2.9K]

Answer: $\sin \theta=\frac{-45}{53}$

Step-by-step explanation:

Since we have given that

$\cos\theta=\frac{28}{53}$

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

$\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}$

It must be negative due to its presence in Fourth quadrant.

Hence, $\sin \theta=\frac{-45}{53}$

7 0

3 years ago