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3 years ago

OA) 30° O B) 60° D) 50° OC) 90°

Mathematics

2 answers:

professor190 [17]3 years ago

5 0

The three inside angles of a triangle need to equal 180 degrees.

You are given the bottom two angles.

75 + 75 = 150

WYZ = 180 - 150 = 30

The answer is 30 degrees.

marissa [1.9K]3 years ago

3 0

Answer:

Step-by-step explanation:

We know that the sum of interior angles in a triangle is 180°. Looking at ΔWYZ, we can write:

? + 75 + 75 = 180

? + 150 = 180

? = 30°

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1) At AJ Welding Company they employ 253 people, 108 employees receive 2 weeks of paid 1) _______ vacation each year. Find the r

Fantom [35]

Answer:

108 : 145

Step-by-step explanation:

253 - 108 = 145

Ratio of those who receive 2 weeks of paid vacation is 108

Ratio of those paid vacation is not 2 weeks is 145

108 : 145

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3 years ago

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position

stira [4]

Answer:

$v(t) = (2t+1) \mathbb{i} +3t^2 \mathbb{j} +4t^3 \mathbb{k}$

$r(t) = (t^2+t) \mathbb{i} +(t^3+2) \mathbb{j} +(t^4 - 3) \mathbb{k}$

Step-by-step explanation:

The velocity vector is the integral of the acceleration vector i.e.

$v(t) = \int a(t) dt$

$v(t) = \int (2 \mathbb{i}+6t \mathbb{j} 12t^2 \mathbb{k}) dt$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + C_1$

When $t=0$ , $v(0) = \mathbb{i}$ . Inserting these values in $v(t)$ ,

$C_1= \mathbb{i}$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + \mathbb{i}$

$v(t) = (2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}$

The position vector is the integral of the velocity vector i.e.

$r(t) = \int v(t) dt$

$r(t) = \int ((2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}) dt$

$r(t) = (t^2+t) \mathbb{i}+t^3 \mathbb{j} t^4 \mathbb{k} + C_2$

When $t=0$ , $r(t) =2\mathbb{j}-3\mathbb{k}$ . Inserting these values in $r(t)$ ,

$C_2=2\mathbb{j}-3\mathbb{k}$

$r(t)= (t^2+t) \mathbb{i}+t^3 \mathbb{j}+ t^4 \mathbb{k} + 2\mathbb{j}-3\mathbb{k}$

$r(t) = (t^2+t) \mathbb{i}+(t^3+2) \mathbb{j}+ (t^4-3) \mathbb{k}$

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4 years ago

What is 2.343 rounded to the nearest tenth

elena-s [515]

I'm pretty sure it's 2.3

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3 years ago

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How would I solve 16x^4-41x^2+25=0 ???

sveticcg [70]

$16{ x }^{ 4 }-41{ x }^{ 2 }+25=0$

${ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0$

First of all to make our equation simpler, we'll equal $x^{2}$ to a variable like 'a'.

So,

${ x }^{ 2 }=a$

Now let's plug $x^{2}$ 's value (a) into the equation.

$16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (<span>fourth-degree <span>equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :
</span></span>
$t{ x }^{ 2 }+bx+c=0$

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

$t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^{ 2 }=a$

So after we found a's solution set, that means.

${ x }^{ 2 }=\frac { 50 }{ 32 }$

and

${ x }^{ 2 }=1$

We'll also solve this equations to find x's solution set :)

${ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }$

${ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

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3 years ago

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Convert 3.8 (8 repeating) into a fraction

navik [9.2K]

Answer:

19/5

Step-by-step explanation:

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