All categories

3 years ago

Find measure of angle 2 - can someone help???

Mathematics

2 answers:

dolphi86 [110]3 years ago

6 0

Answer:

$m\angle 2 = 88\degree - m\angle 1$

Step-by-step explanation:

$m\angle 1 + m\angle 2 = m\angle WUV$

$m\angle 2 = m\angle WUV- m\angle 1$

$m\angle 2 = 88\degree - m\angle 1$

LiRa [457]3 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

A polling organization contacts 2077 adult women who are 30 to 70 years of age and live in the United States and asks whether or

vfiekz [6]

Answer: C. Adult women who are 30 to 70 years of age and live in the United States.

Step-by-step explanation:

The population in the study is the adult women who are 30 to 70 years of age and live in the United States. In statistics, a population is the whole or entire pool of items or event from which a statistical sample is drawn. A population may refer to an entire group of people, objects, events, church visits, food consumed or measurements. In the case above the population from which the sample was drawn is the adult women from age 30 to 70 that lived in the United States, no sample was drawn outside this population.

6 0

3 years ago

This is the question i need help with

vekshin1

I'm pretty sure it should be the 45. or A

4 0

4 years ago

Select True or False for each statement.

SSSSS [86.1K]

$\left( \dfrac 1 {64} \right)^{- 5/6} =64^{5/6} = (\sqrt[6]{64})^5 = 2^5 =32$

TRUE

$\sqrt[5]{36^4}=36^{4/5}$

which surely isn't 36. FALSE

$\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0$

FALSE

For the fourth one we have a

$\sqrt{98b} + \sqrt{2b}$

which isnt

$10\sqrt{b}$

so this is FALSE.

$\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}$

$= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}$

$=(\sqrt 5 + \sqrt 6)^2$

No fractions in that one so FALSE.

3 0

3 years ago

P (2, 3) and Q (6, 10) are two points on the Cartesian plane.

lara31 [8.8K]

Step-by-step explanation:

Given:

$(x_1, \: y_1 ) =(2, \:3)\:\:\&\:\: (x_2, \: y_2) =(6, \:10)$

$l(PQ) = \sqrt{ {(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}} \\ \\ = \sqrt{ {(2 - 6)}^{2} + {(3 - 10)}^{2}} \\ \\ = \sqrt{ {( - 4)}^{2} + {( - 7)}^{2}} \\ \\ = \sqrt{16 + 49} \\ \\ = \sqrt{65} \\ \\ \therefore \purple{ \boxed{ l(PQ) = 8.06 \: units}}\\\\$

Let S be the mid-point of PQ.

$\therefore S = \{\frac{x_1+x_2}{2}, \:\:\frac{y_1+y_2}{2}\} \\\\= \{\frac{2 +6}{2}, \:\:\frac{3+10}{2}\} \\\\= \{\frac{8}{2}, \:\:\frac{13}{2}\} \\\\\therefore S= \{4, \:\:6.5\} \\\\$

Equation of line PQ is given as:

$\frac{y-y_1}{y_1 - y_2}=\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-3}{3 - 10}=\frac{x-2}{2 - 6} \\\\\therefore \frac{y-3}{-7}=\frac{x-2}{-4} \\\\\therefore \frac{y-3}{7}=\frac{x-2}{4} \\\\\therefore 4(y-3)=7(x-2)\\\\\therefore 4y-12=7x-14\\\\\therefore 7x-14 - 4y + 12=0\\\\\red{\boxed {\therefore 7x-4y - 2 =0}}$

4 0

3 years ago

Read 2 more answers

BD bisects ABC. Find the value of x.

yuradex [85]

90-52=38

38=3x-5

3x=43

x=14.3

6 0

2 years ago