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3 years ago

The area of a triangle is found using the following formula:

Mathematics

2 answers:

Elena-2011 [213]3 years ago

8 0

Answer:

12 in. 2

Step-by-step explanation:

3 x 8= 24

1/2. =0.5

24 x 0.5 = 12

Paul [167]3 years ago

3 0

Answer:

12 in.^2

Step-by-step explanation:

The formula to find the area of a triangle is as follows:

1/2bh

When the variables are plugged in, you get:

1/2(3)(8)

First, solve the parentheses. 3*8=24.

Next, divide by 2. 24/2=12.

The answer is 12 inches squared, or 12 in.^2 .

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50 POINTS AND BRAINLIEST! What is the area of this figure?

oksian1 [2.3K]

Answer:

42 units²

Step-by-step explanation:

The figure is composed of a rectangle ( middle section) and 2 triangles with the same length base and height

Area of rectangle = 7 × 4 = 28 units²

Area of 2 triangles = 2 × $\frac{1}{2}$ × 7 × 2 = 14 units²

Area of figure = 28 + 14 = 42 units²

5 0

3 years ago

Read 2 more answers

Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?

garik1379 [7]

Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is $z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$ .

The domain of the function is all real numbers and its range is between -4 and 5.

The graph is enclosed below as attachment.

Step-by-step explanation:

Let be $z (x) = \cos x$ the base formula, where $x$ is measured in sexagesimal degrees. This expression must be transformed by using the following data:

$T = 180^{\circ}$ (Period)

$z_{min} = -4$ (Minimum)

$z_{max} = 5$ (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of $2\pi$ radians. In addition, the following considerations must be taken into account for transformations:

1) $x$ must be replaced by $\frac{2\pi\cdot x}{180^{\circ}}$ . (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

$\Delta z = \frac{z_{max}-z_{min}}{2}$

$\Delta z = \frac{5+4}{2}$

$\Delta z = \frac{9}{2}$

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

$z_{m} = \frac{z_{min}+z_{max}}{2}$

$z_{m} = \frac{1}{2}$

The new function is:

$z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)$

Given that $z_{m} = \frac{1}{2}$ , $\Delta z = \frac{9}{2}$ and $T = 180^{\circ}$ , the outcome is:

$z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

8 0

3 years ago

Solve:<br> 799 \ 4 =<br> What is the answer with the remainder

frozen [14]

The answer of 799/4 is equal 1

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3 years ago

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Find the angle of cd round to the nearest tenth

ololo11 [35]

Answer:
14.1
Step-by-step explanation:

6 0

2 years ago

What types of numbers are undefined when they are under a radical sign? If you were dealing with the number √-1, would it be def

Misha Larkins [42]

The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1

If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1

If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)

If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)

however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1

If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1

and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)

Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.

8 0

2 years ago