Help ASAP PLEASE!!!<br>

Solve the system of equations.

Xelga [282]

Answer:

b. x=1, y=2, z=3

Step-by-step explanation:

The system of equations ...

3x +2y +z = 10
9x -6y +z = 0
x -y -3z = -10

has solution (x, y, z) = (1, 2, 3) . . . . matches choice B.

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While it is convenient to solve this using a graphing calculator or web site, one can easily solve the system by hand.

Subtract the second equation from 3 times the first:

3(3x +2y +z) -(9x -6y +z) = 3(10) -(0)

12y + 2z = 30 . . . . simplify

Dividing this result by 2 gives ...

6y +z = 15 . . . . . . [eq4]

Subtract 3 times the third equation from the first:

(3x +2y +z) -3(x -y -3z) = (10) -3(-10)

5y +10z = 40 . . . . simplify

y + 2z = 8 . . . . . . . divide by 5 . . . . . [eq5]

The two equations [eq4] and [eq5] can be solved any of the ways you usually solve two equations in two variables. Here, we'll use the first equation to write an expression for z that we can substitute into the second equation.

z = 15 -6y . . . . . subtract 6y from [eq4]

y + 2(15 -6y) = 8 . . . . . substitute for z in [eq5]

-11y +30 = 8 . . . . . simplify

-11y = -22 . . . . . . . subtract 30

y = 2 . . . . . . . . . . . divide by the coefficient of y

z = 15 -6(2) = 3 . . . . substitute for y in our equation for z

Substituting these values for y and z into the third original equation gives ...

x - 2 -3(3) = -10

x -11 = -10 . . . . . . . . simplify

x = 1 . . . . . . . . . . . . add 11

The solution to the above system of equations is (x, y, z) = (1, 2, 3).

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<em>Comment on the problem statement</em>

Math is generally unforgiving of imprecision. The given system of equations has no variable "z", and some other typos are apparently involved. That is why we rewrote the system to the equations shown above.

It is very easy to mistake z for 2, or g for 9, or o for 0, or 1 for 7. There are other confusions that are possible, as well. Letters I (eye) and l (ell) are easily confused, and may be confused with 1 (one) as well. Sometimes y and 4, or 4 and 9, can also be written so as to be difficult to tell apart. Great care must be taken when handwriting these symbols.

7 0

4 years ago

How many times does 9 go int 67

AnnZ [28]

°° Awesome question!

I'll show you how to solve this kind of question:

First, you can multiply 9 times any number to see if it's smaller than 67. If it is, it is OK to keep multiplying 9 by more numbers. Let's try this!

9 × 1 obviously equals 9, so we can keep going up.

9 × 2 = 18. Not quite there yet!

Sometimes it is OK for you to skip some numbers, Watch:

9 × 5 = 45. We can keep going. At this point you don't need to go backwards and do 9 × 3 or 9 × 4, because the products will only get farther away from 67.

9 × 6 = 54. We're getting closer!

Let's try to skip one number:

9 × 8 = 72 - HOLD UP! Oh no! We went over a little too much!

Let's try a number in between 6 and 8 - That would be 7, right?

9 × 7 = 63 - We were right on point! Obviously you can't multiply 9 by 8 again because you would get a bigger number - the 72.

__________________________________________________________

That was just one way to go about it, but there's another way, too!

You can divide 9 by 67 and see what you get. You may have a remainder, but you can just throw that out the window when we're all done:

9 ÷ 67 = 7.4444444444444444 The 4 just repeats on and on again, and we don't want that. So just throw those 4's out the window so you're left with 7.

Based on all this work, 9 goes into 67 7 times.

↑ ↑ ↑ Hope this helps! :D

4 0

3 years ago

Read 2 more answers

1. 3,742 x 6= help with step by step

xenn [34]

Answer: The answer is 22,452

Step-by-step explanation: Multiply 6×2 which will be 12

Put the 1 from the 12 on top of the 4 then do 4×6+1 it will equal 25

Then put the 2 on top of the 7 then do 7×6+2 it will equal 44

Then put the 4 on top of the 3 then do 3×6+4 it will equal 22

And that's how you get your answer.

5 0

3 years ago

Suppose sin(A) = 1/4. use the trig identity sin^2(A)+cos^2(A)=1 to find cos(A) in quadrant II. around to ten-thousandth.

larisa86 [58]

In quadrant II, $\cos(A)$ will be negative. So

$\sin^2(A) + \cos^2(A) = 1 \implies \cos(A) = -\sqrt{1 - \sin^2(A)} = -\dfrac{\sqrt{15}}4 \approx \boxed{-0.9682}$

8 0

2 years ago

What is the equation of the circle that has a center at (-3, 1) and a radius of 5? A) (x + 3)2 + (y - 1)2 = 5 B) (x - 3)2 + (y +

Gnoma [55]

Answer:

C

Step-by-step explanation:

The vertex form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$ where (h,k) is the center of the circle and r is the radius. This means that for the circle with center (-3,1) and radius 5 substitute into the vertex form these values. Substitute h = -3, k= 1, and r = 5. The equation becomes $(x--3)^2 + (y-1)^2 = 5^2$ . This simplifies to $(x+3)^2 + (y-1)^2 = 25$ . The solution is C.

7 0

3 years ago

Help ASAP PLEASE!!!​

Help ASAP PLEASE!!!