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3 years ago

-10= -b/4 What Does "b" Equal?

2 answers:

7 0

$-10= \frac{-b}{4}$
Common denominator is 4
$4(-10)= \frac{\not 4(-b)}{\not 4}$
$4(-10)=-b$
$-40=-b$
Multiply by 1
$b=40$
The answer is: b=40

Anarel [89]3 years ago

3 0

$\boxed{-10= \frac{-b}{4} } \ \to -b=-10*4=\boxed{-40} \\ \\ \\\\\huge\text{The answer is -b=-40}$

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PLEASE ANSER URGENT !!!!!

Lynna [10]

The answer is 6/13 since there are 12 distinct letters in the phrase and 12/26 is written as 6/13

3 0

3 years ago

X-5y-z=-20 -2x-y-6z=25 -x-4y-z=-15 Solve by elimination

Tju [1.3M]

Answer:

x = 0

y = 5

z = -5.

Step-by-step explanation:

X-5y-z=-20

-2x-y-6z=25

-x-4y-z=-15

Add equation 1 and 3:

-9y - 2z = -35 .......... A

Multiply the first equation by 2:

2x - 10y - 2z = -40

Now add this to equation 2:

-11y -8z = -15.............B

Now multiply equation A by -4:

36y + 8x = 140.........C

Add equations B and C:

25y = 125

y = 5.

Substitute for y in equation A:

-9*5 - 2z = -35

2z = -45 + 35 = - 10

z = -5.

Now find x by substituting for y and z in the first equation:

x - 5(5) - (-5) = -20

x - 20 = -20

x = 0.

6 0

2 years ago

1. Evaluate: 2 + x - 2(8) for x = 9

Oduvanchick [21]

Answer:

-5

Step-by-step explanation:

2+9-2(8)

11-16

6 0

2 years ago

Read 2 more answers

Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

vodomira [7]

Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two coincident points

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by Pythagorean theorem we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by Zero product property we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0

2 years ago

What is the solution of the following system of linear equations? −21 + 15x = 6y −5x + 2y = −8 Select one: A. (15, -7) B. (4, 11

quester [9]

Answer:

C. No solution

Step-by-step explanation:

System of Equations:

A) $-21+15x=6y$

B) $-5x+2y=-8$

Simplifying and rearranging equation A.

Dividing each term by 3 (common factor for each term) in equation A.

$\frac{-21}{3}+\frac{15x}{3}=\frac{6y}{2}$

$-7+5x=2y$

Subtracting both sides by $2y$

$-7+5x-2y=2y-2y$

$-7+5x-2y=0$

Adding 7 both sides.

$-7+7+5x-2y=0+7$

$5x-2y=7$

Adding the above equation to equation B.

$5x-2y=7$

+ $-5x+2y=-8$

We have $0=-1$

which is not true.

Hence the system has no solution. (Answer)

3 0

2 years ago