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2 years ago

Find the slope of the line that passes through (-1,6) and (15,6)

1 answer:

3 0

Slope = $\frac{Y2 - Y1}{X2 - X1}$
------------------------
6 - 6 = 0
15 - (-1) = 16
0 / 16 = 0
--------------------------------
You have a slope of 0, meaning that this is a straight horizontal line

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Schoolyard has an area of 3,564ft.² the section that is in the shape of a triangle is just for picnic table and what is the area

asambeis [7]

Answer:

The area of the section for the picnic tables is $1,188\ ft^{2}$

Step-by-step explanation:

we know that

The area of the schoolyard is equal to the area of a rectangle plus the area of a triangle

$3,564=44*(2x-x)+\frac{1}{2}(2x-x)*44\\ \\3,564=44x+22x\\\\3,564= 66x\\ \\x=54\ ft$

The area of the triangle is equal to

$22x=22*54=1,188\ ft^{2}$

6 0

2 years ago

Triangle DEF has vertices located at D (2, 1), E (3,5), and F (6,2).

IRINA_888 [86]

Answer:

(A)

Length of DE = √17

EF = √18

DF = √17

(B) Slope of DE = 4

EF = 1

DF = 1/4

Step-by-step explanation:

See attached image.

3 0

2 years ago

(-3) + 8 – 7x7<br> I need help with this on

nika2105 [10]

Answer: -44

Step-by-step explanation: PEMDAS(Parenthesis; Exponent; Multiply; Divide; Add; Subtract)

First you multiply the seven with the other seven to get 49. The problem will look like this: (-3) + 8 - 49. Now just add the -3 and 8. You'd get 5.

Now the problem looks like this: 5 - 49. And you then get -44.

3 0

2 years ago

Fred brought 8 new baseball training cards to add to his collection .the next day his dog ate half of his collection there are n

ollegr [7]

Answer:

27 x 2 = 54

Step-by-step explanation:

3 0

2 years ago

Solve for x, 0 ≤ x ≤2π (2cosx-1)(2sinx+√3 ) = 0

vladimir1956 [14]

Answer:

x = π/3, x = 5π/3, x = 4π/3

Step-by-step explanation:

Let's split the given equation (2cosx-1)(2sinx+√3 ) = 0 into two parts, and solve each separately. These parts would be 2cos(x) - 1 = 0, and 2sin(x) + √3 = 0.

$2\cos \left(x\right)-1=0,\\2\cos \left(x\right)=1,\\\cos \left(x\right)=\frac{1}{2}$

Remember that the general solutions for cos(x) = 1/2 are x = π/3 + 2πn and x = 5π/3 + 2πn. In this case we are given the interval 0 ≤ x ≤2π, and therefore x = π/3, and x = 5π/3.

Similarly:

$\:2\sin \left(x\right)+\sqrt{3}=0,\\2\sin \left(x\right)=-\sqrt{3},\\\sin \left(x\right)=-\frac{\sqrt{3}}{2}$

The general solutions for sin(x) = - √3/2 are x = 4π/3 + 2πn and x = 5π/3 + 2πn. Therefore x = 4π/3 and x = 5π/3 in this case.

So we have x = π/3, x = 5π/3, and x = 4π/3 as our solutions.

5 0

3 years ago