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3 years ago

What is the y-value in the solution to this system of linear equations?

Mathematics

2 answers:

baherus [9]3 years ago

4 0

Answer:

Option A.)−4

Step-by-step explanation:

we have

4x+5y=-12 -----> equation A

-2x+3y=-16 -----> equation B

Solve the system of equations by elimination

Multiply the equation B by 2

2*(-2x+3y)=-16*2

-4x+6y=-32 -----> equation C

Adds equation A and equation C

4x+5y=-12

-4x+6y=-32

-----------------

5y+6y=-12-32

11y=-44

y=-4

Alik [6]3 years ago

3 0

Answer with Step-by-step explanation:

We are given a system of linear equations:

4x + 5y = −12

-2x + 3y = −16

Multiply second equation by 2 and add it to first equation.

4x+5y-4x+6y= -12-32

⇒ 11y= -44

Dividing both sides by 11

⇒ y= -4

Hence, the y-value in the solution to the system of linear equations

4x + 5y = −12

-2x + 3y = −16 is:

A.)−4

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drawbridge at the entrance to an ancient castle is raised and lowered by a pair of chains. The figure represents the drawbridge

jasenka [17]

Answer:

4.0 meters, ∠C = 39°, ∠A = 51°

Step-by-step explanation:

Firstly, our diagram shows us that the given triangle is actually a right triangle. So we can use the Pythagorean Theorem to solve for the height of the chain:

$a^{2} +b^{2} =c^{2}$

$(3.3)^{2} +b^{2} =(5.2)^{2}$

$b^{2} =(5.2)^{2}-(3.3)^{2}$

$b =\sqrt{(5.2)^{2}-(3.3)^{2}}$

$b=4.0187...$

$b=4.0 m$

Now, we can use the Law of Cosines to figure out one of the acute angles:

$c^{2} =a^{2} +b^{2} -2ab(cos(C))$

$(3.3)^{2} =(4.0)^{2} +(5.2)^{2} -2(4.0)(5.2)(cos(C))$

$cos(C)=\frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)}$

$C=cos^{-1}( \frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)})$

$C=39.3915...$

∠C = 39°

And since we know that all angles in a triangle add up to 180°:

∠A + ∠B + ∠C = 180

∠A + 90 + 39 = 180

∠A = 180 - 90 - 39

∠A = 51°

However, you should always review any answers on the Internet and make sure they are correct! Check my work to see if I made any mistakes!

7 0

3 years ago

An outside thermometer reads -5 degrees at 8:00 a.m. By 12:30 p.m. the temperature outside has increased by 11 degrees. At 12:30

Lesechka [4]

Answer:

73 degrees.

Step-by-step explanation:

So, let's see the information provided in the question.

At 8 AM, the thermometer was reading -5 degrees.

At 12:30 PM, the temperature has increased by 11 degrees, so it's now +6 degrees outside.

Inside Susan's house, the temperature is 7 degrees more than 11 times the outside temperature, at 12:30 PM.

So, we know the outside temperature at 12:30 PM is +6 degrees.

11 times 6 = 66 degrees.

Then we add the additional 7 degrees, to get a total of 73 degrees.

7 0

3 years ago

Make a table showing the probability distribution for the possible sums when tossing two four-sided dice (the sides are numbered

alexira [117]

We would have the following sample space:
(1, 1), (1, 2), (1, 3), (1, 4)
(2, 1), (2, 2), (2, 3), (2, 4)
(3, 1), (3, 2), (3, 3), (3, 4)
(4, 1), (4, 2), (4, 3), (4, 4)

Those give us these sums:
2, 3, 4, 5
3, 4, 5, 6
4, 5, 6, 7
5, 6, 7, 8

P(sum of 2) = 1/16 =0.0625
P(sum of 3) = 2/16 = 0.125
P(sum of 4) = 3/16 = 0.1875
P(sum of 5) = 4/16 = 0.25
P(sum of 6) = 3/16 = 0.1875
P(sum of 7) = 2/16 = 0.125
P(sum of 8) = 1/16 = 0.0625

8 0

4 years ago

Part 4: Identify the vertex, focus and directrix of each. Then sketch the graph.

Nezavi [6.7K]

Answer: 1) Vertex: (6, -2) Focus: (6, -7/4) Directrix: y = -9/4

2) Vertex: (-2, -1) Focus: (-7/4, -1) Directrix: x = -9/4

Step-by-step explanation:

Rewrite the equation in vertex format y = a(x - h)² + k or x = a(y - k)² + h by completing the square. Divide the b-value by 2 and square it - add that value to both sides of the equation.

(h, k) is the vertex
p is the distance from the vertex to the focus
-p is the distance from the vertex to the directrix

$\bullet\quad a=\dfrac{1}{4p}$

1) y = x² - 12x + 34

$y-34=x^2-12x\\\\y-34+\bigg(\dfrac{-12}{2}\bigg)^2=x^2-12x+\bigg(\dfrac{-12}{2}\bigg)^2\\\\\\y-34+36=(x-6)^2\\\\y+2=(x-6)^2\\\\y=(x-6)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(6,-2)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(6,-\dfrac{7}{4}\bigg)\\\\\\\text{Directrix: y= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad y=-\dfrac{9}{4}$

*******************************************************************************************

2) x = y² + 2y - 1

$x+1=y^2+2y\\\\x+1+\bigg(\dfrac{2}{2}\bigg)^2=y^2+2y+\bigg(\dfrac{2}{2}\bigg)^2\\\\\\x+1+1=(y+1)^2\\\\x+2=(y+1)^2\\\\x=(y+1)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(-2,-1)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(-\dfrac{7}{4},-1\bigg)\\\\\\\text{Directrix: x= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=-\dfrac{9}{4}$

4 0

3 years ago

Can someone please help me with this???

tino4ka555 [31]

Answer:

1 cm

Step-by-step explanation:

To solve this problem we can use the Pythagorean theorem. In fact the diagonal of a rectangle is an hypotenuse of a right triangle, while the length is a leg. The width is the other leg

width = √2^2 - (√3)^2 = √4 - 3 = √1 = 1 cm

8 0

3 years ago