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Snezhnost [94]
3 years ago
6

What is the area of a triangle with vertices at (−4, 1) , ​ (−7, 5) ​ , and ​ (0, 1) ​ ?

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
3 0
Plot the points on a graph. Connect the dots into a triangle. See that the height of the triangle is from y=5 down to y=1. So the height is 4 units.
 Area of a triangle: A = (1/2)BH

You need to find the length of the base. Which is from point (-4,1) to (0,1). You can use the distance formula r just see from the graph that the base is 4 units.

A = (1/2)(4)(4)
A = 8

** Distance formula fyi
d² = (X-x)² + (Y-y)²
with points (X,Y) and (x,y)
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Please help!
Marat540 [252]

Answer: width = 82.50 m

Step-by-step explanation:

P = 2l x 2w

425 = 2(130) + 2w

425 - 260 = 2w

w = 165 / 2 = 82.50

5 0
3 years ago
- 18 = -3x + 6<br> Plz help
exis [7]

Answer:

8 =x

Step-by-step explanation:

- 18 = -3x + 6

Subtract 6 from each side

-18-6 = -3x+6-6

-24 = -3x

Divide each side by -3

-24/-3 = -3x/-3

8 =x

8 0
3 years ago
Read 2 more answers
What percent is equal to 7/25​
ella [17]
28% because 25x4=100 7x4=28
8 0
3 years ago
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e−5t
Elan Coil [88]

Answer:

x = 1 - 5t

y = t

z = 1 - 5t

Step-by-step explanation:

For the equation of a line, we need a point and a direction vector. We are given a point (1, 0, 1).

Since the line is suppose to be a tangent to the given curve at the point (1, 0, 1), we need to find a tangent vector for which the curve passes through that point.

We have x = e^(-5t)cos5t

at t = 1, x = e^(-5)cos5

at t = 0, x = 1

y = e^(-5t)sin5t

at t = 1, y = e^(-5)sin5

at t = 0, y = 0

z = e^(-5t)

at t = 1, z = e^(-5)

at t = 0, z = 1

Clearly, the only parameter value for which the curve passes through the point (1, 0, 1) is t = 0.

In vector notation, the curve

r(t) = xi + yj + zk

= e^(-5t)cos5t i + e^(-5t)sin5t j + e^(-5t) k

r'(t) = [-5e^(-5t)cos5t - e^(-5t)sin5t] i +[e^(-5t)cos5t - 5e^(-5t)sin5t] j - 5e^(-5t) k

r'(0) = -5i + j - 5k

is a vector tangent at the point.

We get the parametric equation from this.

x = x(0) + tx'(0)

= 1 - 5t

y = y(0) + ty'(0)

= t

z = z(0) + tz'(0)

= 1 - 5t

8 0
3 years ago
The star means that is my guess.
OLEGan [10]
You are correct for all of them,
Precision goes up as you have more decimals. It gives you a more accurate measurement regardless of what it is measuring.
A line graph measures two variables, the X and the Y, and how they change based off of one another
And an educated guess would usually based on previous trends or results
7 0
3 years ago
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