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3 years ago

In the equations of Law of Cosines do I multiply the two by both a AND b or just A?

Mathematics

1 answer:

SashulF [63]3 years ago

5 0

The final answer is 34.0

See the attached image to see how I got that answer.

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Help me please I don’t know how to

Irina-Kira [14]

Answer:

A: 22

Step-by-step explanation:

2The interquartile range begins at 45 and ends at 67. All you need to do is subtract 45 drom 67, and you get 22.

4 0

3 years ago

Which equation represents a line which is parallel to the line 8y-3x=-8

marin [14]

First, we need to make the equation given to be in the form;

y=mx + b

where m is the slope and b is the intercept

8y - 3x = -8

Add 2x to both-side of the equation

8y = 3x - 8

Divide through the equation by 8

y = 3/8 x -1

comparing the above with y =mx + b

m = 3/8

slope of parallel equations are equal

From the options given the only equation with slope 3/8 is;

y= 3/8 x + 8

$y=\frac{3}{8}x\text{ + 8}$

The above equation of the line is parallel to 8y - 3x = -8

7 0

1 year ago

What is the answer to -6+7x-6x=3

fgiga [73]

Hi!

-6+7x-6x = 3
-6+x = 3
x = 3+6
x = 9

Answer:

x = 9

3 0

3 years ago

What things should be written in the conclusion of letter <br>

GarryVolchara [31]

Answer:

Your signature, and a nice 4-word goodbye

Step-by-step explanation:

Your signature so they know it was YOU who wrote it and a nice 4-word goodbye to show the other person that you care.

3 0

3 years ago

Find the zeros of y = x2 – 6x – 4 by completing the square.

katrin [286]

Answer:

The solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

Step-by-step explanation:

Given the function

$y\:=\:x^2\:-\:6x\:-\:4$

substitute y = 0 in the equation to determine the zeros

$0\:=\:x^2\:-\:6x\:-\:4$

Switch sides

$x^2-6x-4=0$

Add 4 to both sides

$x^2-6x-4+4=0+4$

Simplify

$x^2-6x=4$

Rewrite in the form (x+a)² = b

But, in order to rewrite in the form x²+2ax+a²

Solve for 'a'

2ax = -6x

a = -3

so add a² = (-3)² to both sides

$x^2-6x+\left(-3\right)^2=4+\left(-3\right)^2$

$x^2-6x+\left(-3\right)^2=13$

Apply perfect square formula: (a-b)² = a²-2ab+b²

$\left(x-3\right)^2=13$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-3=\sqrt{13}$

Add 3 to both sides

$x-3+3=\sqrt{13}+3$

Simplify

$x=\sqrt{13}+3$

now solving

$x-3=-\sqrt{13}$

Add 3 to both sides

$x-3+3=-\sqrt{13}+3$

Simplify

$x=-\sqrt{13}+3$

Thus, the solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

4 0

3 years ago