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3 years ago

The area, A, of a triangle is one-half the product of the triangle's base and height.

Mathematics

1 answer:

nordsb [41]3 years ago

7 0

A = (1/2)(b * h)

It is 1/2 times the product (which is the end result of multiplication of two values), of base (b) times height (h)

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Mr. Campbell. paid $48 for 12 cartons of juice.

Wewaii [24]

Answer:

Step-by-step explanation:

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3 years ago

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I need help in an hour or less pls The number N of bacteria in a refrigerated food is given by N(T ) = 20T 2 – 80T + 500, 2 ≤ T

weqwewe [10]

Answer:

The number N of bacteria in a refrigerated food is given by N(T) = 10T^2 - 20T + 600, 2 ≤ T ≤ 20 where T is the temperature of the food in degrees Celsius. When the food is removed from refrigeration, the temperature of the food is given by T(t) = 3t + 2, 0 ≤ t ≤ 6 where t is the time in hours.

Step-by-step explanation:

7 0

3 years ago

If AC is parallel to DE, find X

UNO [17]

Answer: x = 50

Concept:

Here, we need to know the idea of alternative interior angles and the angle sum theorem.

Alternative interior angles are angles that are formed inside the two parallel lines, and the values are equal.

The angle sum theorem implies that the sum of interior angles of a triangle is 180°

If you are still confused, please refer to the attachment below or let me know.

Step-by-step explanation:

Given information:

AC ║ DE

∠ABC = 85°

∠A = 135°

Find the value of ∠BAC

∠A + ∠BAC = 180° (Supplementary angle)

(135°) + ∠BAC = 180°

∠BAC = 45°

Find the value of ∠BCA

∠ABC + ∠BAC + ∠BCA = 180° (Angle sum theorem)

(85°) + (45°) + ∠BCA = 180°

∠BCA = 50°

Find the value of x (∠EBC)

∠EBC ≅ ∠BCA (Alternative interior angles)

Since, ∠BCA = 50°

Therefore, ∠EBC = 50°

$\Large\boxed{x=50}$

Hope this helps!! :)

Please let me know if you have any questions

8 0

1 year ago

Convert the rectangular coordinates (-9, 3V3) into polar form. Express the angle

Whitepunk [10]

Answer:

$(6\sqrt{3},\,\frac{5\pi}{6})$

Step-by-step explanation:

The radius r can be found from the relationship

$r^2=x^2+y^2\\r^2=(-9)^2+(3\sqrt{3})^2\\r^2=81+27=108\\r=\sqrt{108}\\r=6\sqrt{3}$

The point is in Quadrant II (-, +), so use the inverse cosine function to find the angle.

$\cos{\theta}=\frac{x}{r}=\frac{-9}{6\sqrt{3}}\\\cos{\theta}=-\frac{9}{6\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\\cos{\theta}=-\frac{9\sqrt{3}}{6\cdot3}\\\cos{\theta}=-\frac{\sqrt{3}}{2}\\\\\cos^{-1}\frac{-\sqrt{3}}{2}}=\frac{5\pi}{6}$