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3 years ago

1. Solve the system of equations algebraically.

Mathematics

2 answers:

goldfiish [28.3K]3 years ago

8 0

Hello,
Please, see the attached files.
Thanks.

Andrew [12]3 years ago

4 0

1. We can double the second equation to get: 2x - 6y = 6; and then add the equations together to get -y = 4, y = -4. We plug y in to get x - 3(-4) = 3, x = 15. The answer would be (15, -4)
2. We can subtract the two equations to eliminate y and get 2x = 12, x = 6. We can plug x in to the second equation to get 9 + 3y = 18, and 3y = 9, so y = 3. The answer would be (6, 3)
3. We add the first two equations to get 8x - 5y = -34, and then subtract the last two equations to get 3x - 2y = -13. We multiply the first by 2, and the second by 5, and then add them together. We get x = -3. We plug the x-value into the second equation to get -9 - 2y = -13, so y = 2. We plug both values into the third equation to get -9 - 2 - z = -6. -z = 5, z = -5.
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Hope this helps!
==jding713==

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Given: AB = 4 AC = 6 What point is in the interior of both circles? H. A. B.

AleksAgata [21]

Answer:

Point A is in the interior of both circles

Step-by-step explanation:

Verify each point

we know that

Point H is in the interior of greater circle but is outside smaller circle

Point A is the center both circles, then is in the interior of greater circle and is in the interior of smaller circle

Point B is on the smaller circle and is in the interior of greater circle

therefore

Point A is in the interior of both circles

5 0

4 years ago

At the end of the month mr copley has $1473.61 in is checking account his checkbook showed the following transaction if made mad

ycow [4]

Total balance in Mr. Copley at the end of the month = $1473.61.

Initial deposite = $75 (not shown in his checkbook).

Let us assume his balance at the beginning of the month = x.

Total balance at the end of the month = The balance at the beginning of the month + total deposite.

$1473.61 = x + 75.

Subtracting 75 from both sides, we get

1473.61 -75 = x + 75 -75.

1398.61 = x.

Therefore, his balance at the beginning of the month was $1398.61.

7 0

4 years ago

CALC- limits please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago

There are 45 Vehicles in the in a parking lot 3/5 of the vehicles are minivan how many vehicles in the parking lot at minivans

taurus [48]

27 vehicles

3/5 x 45
5 goes into 45 9 times, 9x3= 27, therefore there are 27 minivans

7 0

3 years ago

Please Help ME!!!

Airida [17]

Answer:

Hence Group A compose of the alternate Addition of "ONE" and "THREE" units

Hence Group B Composes of alternate 22 units addition and 3 units addition.

Step-by-step explanation:

Given:

Two groups as A and B with number in certain patterns as

To Find:

What is that certain pattern in numbers?

Solution:

Group A includes the following numbers as ,

0,1,4,5,8

Here 1st number=0 then 2nd number =1

So in between this addition is 1 is present

And then 3rd number =4 and 4th number=5

so there alternate addition of "one"(1) present

Now in 3rd and 2nd number there is addition of 3 units

And in 5th and 4 number there is addition of 3 units .

Hence this group compose of the alternate Addition of "ONE" and "THREE" units.

Group B includes the following number as ,

10,32,35,57,79

So here 1st number is 10 and 2nd number is 32 (Addition of 22 units).

Further alternate 4th 57 and 3rd number is 35 (addition of 22 units).

For 3rd number is 35 and 2nd number is 32 (addition of 3 units)

For 6th number will be is 82 and 5th number is 79 (addition of 3 units).

Hence this group Composes of alternate 22 units addition and 3 units addition.

3 0

3 years ago