All categories

3 years ago

In a trapezoid, which two segments will always be parallel?

Mathematics

2 answers:

AleksAgata [21]3 years ago

8 0

Answer:

Step-by-step explanation:

Alchen [17]3 years ago

5 0

The two parallel lines of a trapezoid are called the bases, it’s d

You might be interested in

The area of this rectangle is 165 sq meters the width is 22 meters what’s is the perimeter?

Vesna [10]

Answer:

59 meters

Step-by-step explanation:

A=L*W

165=L*22

L=165/22=7.5

7.5=L

P=L+L+W+W

7.5+7.5+22+22=59

7 0

3 years ago

Evaluate a + b to the 2nd power for a = 2 and b = 3

jarptica [38.1K]

Answer:

Step-by-step explanation:

You add 2+3 and get 5. Then you square it to get 25.

4 0

3 years ago

Read 2 more answers

The tests done to determine if someone is HIV positive are called Enzyme immunoassay or

Blizzard [7]

The answer would be answer choice B.

8 0

3 years ago

Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits

scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

$x^{2} + 3x - 8^{- 14} = 0$

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

The above solution is for the quadratic equation of the form:

$ax^{2} + bx + c = 0$

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

From the given eqn

a = 1

b = 3

c = $- 8^{- 14}$

Now, using the above values in the formula mentioned above:

$x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)$

Now, Rationalizing the above eqn:

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

$x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

Solving the above eqn:

$x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}$

Solving with the help of caculator:

$x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}$

The precise value upto three decimal places comes out to be:

$x_{1, 1'} = 0.758\times 10^{- 14}$

5 0

3 years ago

Write a linear function f with f(−9)=10 and f(−1)=−2

pashok25 [27]

First we need to find the slope of the function

The slope of a function is equal to

$\frac{f(x_2)-f(x_1)}{x_2-x_1}$

So now we plug this into the equation and find the slope (M will represent the slope)

$m=\frac{10--2}{-9--1}\\m=\frac{12}{-8}\\m=-\frac{3}{2}$

So the slope is -3/2

Now we can use point-slope form.

Point slope form is represented by

$Y-Y_1=m(X-X_1)$

So when we plug in our values we get

$Y--2=-3/2(X--1)\\Y+2 = -3/2(X+1)\\Y+2 = -3/2X-3/2\\Y=-3/2X-7/2$

So the equation is

Y = -3/2X-7/2

6 0

3 years ago