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never [62]
3 years ago
9

How to find area of 2d triangle

Mathematics
1 answer:
amm18123 years ago
3 0

Answer:

Multiply both sides, then divide by 2.

Step-by-step explanation:

A triangle is either half of a square, or half of a rectangle.  The two numbers given are the dimensions to the quadrilateral.  Halving it gives you the area of a triangle.

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ITS URGENT
Vinvika [58]

log.8 (.4096) =4

.8 ^4 = .4096



loga b =c

a^c =b

8 0
3 years ago
point C (4, -3) is translated left 3 and up 5 and is then dilated by a scale factor of 4. What are the coordinates of C'?
sergeinik [125]

Answer: (4, 8)

Step-by-step explanation:

When you translate something to the left you need to subtract the amount you moved left from the original x input

1. Subtract 3 from the coordinates x input

4 - 3 = 1

When you translate something up you need to add the amount you moved up to the original y output

2. Add 5 to the coordinates y output

-3 + 5 = 2

When done translating to the left and up your coordinate is now (1, 2)

When dialating something you need to multiply the scale factor in which you dialated the coordinate to both the x input and y output

3. Multiply 4 to the x input and y output of your newley translated coordinate

1 × 4 = 4 (x input)

2 × 4 = 8 (y output)

So now your final coordinate for C' is (4, 8)!

Hope this helps :)

8 0
3 years ago
Select all the correct answers.
IgorC [24]
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3 0
3 years ago
Read 2 more answers
The intersection of plane R and plane P is
Paraphin [41]

Answer: Line DB

Think of the two planes as pages in a book. They are joined by the spine, which is where the two planes intersect. In this case, that intersection is the line DB.

Point D is on both planes R and S. So is point B. Every point on line DB is found on both planes.

4 0
4 years ago
Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the give
neonofarm [45]

Answer:

Equation of tangent:

y = xe^{x}

At point (1,0):

y = 2.713

Step-by-step explanation:

The equation of tangent line to the function can be calculated by taking the first derivative.

We have,

y = xe^{x}-e^{x}\\\frac{dy}{dx}=\frac{d}{dx}[ xe^{x} ]-\frac{d}{dx} [e^{x}]\\

Applying Product Rule:

d/dx [u.v] = (d/dx u) . (v) + (u) . (d/dx v)

Therefore,

\frac{dy}{dx}=\frac{d}{dx}(x) . e^{x}+x .\frac{d}{dx}(e^{x})- \frac{d}{dx}(e^{x})\\\\\frac{dy}{dx}=(1)(e^{x})+(x)(e^{x})-e^{x}\\ \frac{dy}{dx}=xe^{x}\\\\

The above equation is the equation of tangent line.

The point given is (1,0):

So,

\frac{dy}{dx} = (1)e^{1}\\ \frac{dy}{dx} = e\\\frac{dy}{dx} = 2.713

3 0
3 years ago
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