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2 years ago

11

Point E is located at (2, −3), and point F is located at (−2, −1). Find the point that is 3 over 4 the distance from point E to

point F.

1 answer:

Alisiya [41]2 years ago

6 0

Your answer will be -2+2+4= 4
-3-1+4=0
so it will be 4,0

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Please help me with this. thank you!

Firlakuza [10]

Answer:

1. y = mx + b

Step-by-step explanation:

y = mx + b

2) to find the y intercept you need to see what cross | ( 8,5)

3) 5 = m(8) = b

4) your slope is negative form the why it pointing

5) in this case your slope is 1/3

6) so now you can do your first step now that you have your (y,x) and slope (your solving for (b)

7)

5 = 1/3 (8) + b

5 = 2.6 + b

5 = 2.6

-2.6

2.4 = b

6 0

1 year ago

Read 2 more answers

Given that m 2 = 56 in the diagram what is m 8<br>A 134<br>B 66<br>C 124 <br>D 56

algol [13]

The answer is:
Letter B

5 0

2 years ago

Evaluate the definite integral using the graph of f(x)<br> (Image included)

Tanya [424]

a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

$\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}$

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

$\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}$

c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so

$\displaystyle \int_5^9 f(x) \, dx = -4\times5 = -20$

Then by linearity, we have

$\displaystyle \int_0^9 f(x) \, dx = \left\{\int_0^3 + \int_3^5 + \int_5^9\right\} f(x) \, dx = \frac{15}2 + 0 - 20 = \boxed{-\frac{25}2}$

8 0

1 year ago

The electric heaters, in a 240 volt electric furnace, draw 40 amps of current. What is the total KW rating of the heaters?

expeople1 [14]

9514 1404 393

Answer:

9.6 kW

Step-by-step explanation:

The product of volts and amps is watts. Here, that product is ...

(240 V)(40 A) = 9600 W

1000 W is 1 kW, so this quantity is 9.6 kW

__

This tells the power consumption. The "rating" may be different.

3 0

2 years ago

From a point A that is 8.20 m above level ground, the angle of elevation of the top of a building s 31 deg 20 min and the angle

Mariulka [41]

Answer:

30.11 meters ( approx )

Step-by-step explanation:

Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

Point A is 8.20 m above level ground,

So, the height of the building = ( x + 8.20 ) meters,

Now, 1 degree = 60 minutes,

⇒ $1\text{ minute } =\frac{1}{60}\text{ degree }$

$20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}$

$50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}$

By the below diagram,

$tan ( 12^{\circ} 50') = \frac{8.20}{y}$

$tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}$

$tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}$

$\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}$

Now, again by the below diagram,

$tan (31^{\circ}20')=\frac{x}{y}$

$tan(31+\frac{1}{3})=\frac{x}{y}$

$\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91$

Hence, the height of the building = x + 8.20 = 30.11 meters (approx)

8 0

2 years ago