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3 years ago

11

Two lines are intersected by a third line. If 2 6, which must be true about 2? 2 5 2 is complementary to 5. m2 = m8 2 is supplem

entary to 8.

2 answers:

djverab [1.8K]3 years ago

7 0

Answer:

Angle2 is supplementary to Angle8.

Step-by-step explanation:

sineoko [7]3 years ago

4 0

If this scenario is a pair of parallel lines intersected by a transversal then it is true that
2 = 6
Then, among the choices, the statement that is true is
2 is supplementary to 8 since 6 is supplementary to 8
By transitivity property, 2 is supplementary 8

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What is the General form of the parabola of (y-3)^2=6(x+8)

MatroZZZ [7]

Answer:

x = 1.5 $t^{2}$ , y = 3 + 2at.

Step-by-step explanation:

For the parabola $Y^{2}$ = 4aX ,

General form will be

(X = a $t^{2}$ , Y = 2at) ,

Thus , for the parabola $(y-3)^{2}$ = 6(x +8)

Here , a = 1.5 and Y from the above equation should be substituted by y - 3 and X must be substituted by x + 8. After substitution of the same we can use the general equation formula for this parabola also.

Thus , general equation comes out to be :-

x + 8 = 1.5 $t^{2}$ , y - 3 = 2at

x = 1.5 $t^{2}$ , y = 3 + 2at.

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3 years ago

Read 2 more answers

When the prefix milli is used, it means multiply the number by ? ?

valina [46]

1 meter = 1000 millimeters

so, 5 meters = 5*1000 = 5000 millimeters

and 6000 millimeters = 6000 / 1000 = 6 meters

3 0

3 years ago

M angle 3= 31°. Find m angle 4 (complementary angle)

solniwko [45]

Hi, there.

________

$\boxed{\begin{minipage}{8cm} Complementary angles \\ Complementary angles are angles whose sum \\ is 90^\circ. \text{Using this info, let's consider this:} \\ \text{If an angle's measure is 31 degrees, what's} \\ \text{its \ complement?}\end{minipage}}$

$\text{Set up an equation, letting a be the unknown angle}$

$\bold{a+31=90}\\\text{Subtract 31 from both sides}\\\bold{a=59}$

Hope the answer - and explanation - made sense,

happy studying!!

6 0

2 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

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3 years ago