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3 years ago

6

and what are the coordinates of the first car? after the rotation of 270 degrees counterclockwise about the origin? The coordina

tes of the first car are?? After a rotation of 270 degrees about the origin

1 answer:

monitta3 years ago

6 0

Do you have a picture of the problem to go with this because its not enough info to solve it

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Mr. Skinner has four dogs. He recorded the number of cups of dog food each dog ate over the same month.

GarryVolchara [31]

So to find the answers for each of these, divide the first dog's amount mentioned by the second dog's amount mentioned. For instance, the first one would be 45 over 60, also equal to 3/4. So Bailey ate 3/4 cup for every cup fido ate. Repeat this same process with all the other questions and you can get the following answers:
Fido ate 1 1/3 cup for each cup Bailey ate.
Gunner ate 2 1/2 cups for each cup Spot ate.
Spot ate 2/5 cup for each cup Gunner ate.
Hope this helps!

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3 years ago

Convert 9354grams to kilograms

STatiana [176]

Answer:

9.354kg

Step-by-step explanation:

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3 years ago

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Natali5045456 [20]

The answer would be A

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4 years ago

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julia-pushkina [17]

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3 0

3 years ago

Given: ∆ABC, m∠C = 90° CB = 8, m∠B = 38º Find the area of a circumscribed circle. Find the area of the inscribed circle.

vitfil [10]

Answer:

Circumscribed circle: Around 80.95

Inscribed circle: Around 3.298

Step-by-step explanation:

Since C is a right angle, when the circle is circumscribed it will be an inscribed angle with a corresponding arc length of 2*90=180 degrees. This means that AB is the diameter of the circle. Since the cosine of an angle in a right triangle is equivalent to the length of the adjacent side divided by the length of the hypotenuse:

$\cos 38= \dfrac{8}{AB} \\\\\\AB=\dfrac{8}{\cos 38}\approx 10.152$

To find the area of the circumscribed circle:

$r=\dfrac{AB}{2}\approx 5.076 \\\\\\A=\pi r^2\approx 80.95$

To find the area of the inscribed circle, you need the length of AC, which you can find with the Pythagorean Theorem:

$AC=\sqrt{10.152^2-8^2}\approx 6.25$

The area of the triangle is:

$A=\dfrac{bh}{2}=\dfrac{8\cdot 6.25}{2}=25$

The semiperimeter of the triangle is:

$\dfrac{10.152+6.25+8}{2}\approx 24.4$

The radius of the circle is therefore $\dfrac{25}{24.4}\approx 1.025$

The area of the inscribed circle then is $\pi\cdot (1.025)^2\approx 3.298$ .

Hope this helps!

6 0

4 years ago