\dfrac{1}{3}(4\cdot3)+2^3= 3 1 (4⋅3)+2 3

PLEASE ANSWER ASAP!!! SHOW ALL THE STEPS.

fenix001 [56]

Answer:

For the first question $x = 4\\ y = -1$ and

For the second question $x = 0.5\\ y = -1$

Step-by-step explanation:

Given:

1.

x - 3y = 7

3x + 3y = 9

2.

8x+ 3y = 1

4x + 2y = 0

Elimination method :

In the elimination method we need to make the coefficient of x or the coefficient of y same in both the equation so by adding or subtracting we can eliminate the x term or the y term.

Then substitute that values which you will get on eliminating in any equation you will get the corresponding value.

For the first question, the y coefficient is same hence by adding both the equation we can eliminate 3y term. so on solving we get

$(x - 3y) + (3x + 3y) = 7 + 9\\4x = 16\\x = \frac{16}{4}\\x = 4$

Now substitute X equal to 4 in equation x -3y = 7 we get

$4 - 3y = 7\\-3y = 7 - 4\\-3y = 3\\y = \frac{3}{-3}\\ y = -1\\$

This way we have x is equal to 4 and y is equal to -1 for question number 1.

For the second question, we will make X coefficient same in the second equation that is multiplying by 2 to the equation 4x + 2y = 0 then we get

$8x + 4y = 0\\$

Now the coefficient of x term become same now we will subtract the two equations that is 8x + 3y = 1 and 8x + 4y =0 we get

$(8x + 3y) - (8x + 4y) = 1 - 0\\3y - 4y = 1\\ -y = 1\\y = -1$

Now substitute y equal to -1 in equation 8x +3y = 1 we get

$8x + 3\times -1 = 1\\8x - 3 = 1\\8x = 1 + 3\\8x = 4\\x = \frac{4}{8}\\ x = 0.5\\$

This way we have x is equal to 0.5 and y is equal to -1 for question number 2.

3 0

3 years ago

In need of some help! Thank you in advance. 100 points <3

Anni [7]

Answer:

1

the equation (x-4)^2-17=8 has two solutions because it is in a quadratic form.

(x-4)^2-17=8

x²-8x+16-17-8=0

x²-8x-9=0

Doing middle term factorization

x²-9x+x-9=0

x(x-9)+1(x-9)=0

(x-9)(x+1)=0

either

x=9

x=9or

x=9orx=-1

2.

x^2+4x+3=0:

Doing middle term factorization

x²+3x+x+3=0

x(x+3)+1(x+3)=0

(x+3)(x+1)=0

either

x=-3

or

x=-1

Josh’s solution is incorrect because he missed x=-3.

3.

x^2-6x+7=0

By using quadratic equation formula:

Comparing above equation with ax²+bx+c=0

we get

a=1

b=-6

c=7

Now

we have

x= $\frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}$

x= $\frac{ 6± \sqrt{ {-6}^{2} - 4*1*7} }{2}$

x= $\frac{ 6±2\sqrt{ 2}}{2}$

x= ${ 3±\sqrt{ 2}}$

Taking positive

x= ${ 3+\sqrt{ 2}}$

taking negative

x= ${ 3-\sqrt{ 2}}$

4 0

3 years ago

Using the box-and-whisker plot shown, find the quartile values Q1 and Q3

astra-53 [7]

Answer:

see explanation

Step-by-step explanation:

The lower quartile $Q_{1}$ is positioned at the left side of the box, that is

$Q_{1}$ = - 4

The upper quartile $Q_{3}$ is positioned at the right side of the box

$Q_{3}$ = 6

7 0

4 years ago

A landscaper has enough cement to make a patio with an area of 150 Sq ft. The homeowner wants the length of the patio to be 6 ft

Natasha_Volkova [10]

$A=lw=150\\l=w+6\\(w+6)w=150\\w^2+6w=150\\w^2+6w-150=0\\w=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\w=\frac{-(6)+-\sqrt{(6)^2-4(1)(-150)}}{2(1)}\\w=\frac{-6+-\sqrt{36+600}}{2}\\w=\frac{-6+-\sqrt{636}}{2}\\w=\frac{-6+-2\sqrt{159}}{2}\\w=-3+-\sqrt{159}>0\\w=-3+\sqrt{159}\\\\l=(-3+\sqrt{159})+6\\l=3+\sqrt{159}$

Width = -3 + sqrt(159)
Length = 3 + sqrt(159)

4 0

3 years ago

Los angles is about 385 miles from san francisco. how far apart should the city's be on a map with a scale of 1in=1515mil

klasskru [66]

1 inch=1515 miles
X inches= 385 miles
so 1515 is 3.935 times bigger than 385 miles.
meaning that 1 inch is also 3.935 times bigger than X inches.
to find X we just divide 1 by 3.935
which is: 0.254 inches

4 0

3 years ago

\dfrac{1}{3}(4\cdot3)+2^3= 3 1 ​ (4⋅3)+2 3

\dfrac{1}{3}(4\cdot3)+2^3= 3 1 (4⋅3)+2 3