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AnnZ [28]
3 years ago
9

How am I supposed to solve this? I know how to do the equations but I don't know why there are only 2 places to put an answer fo

r X and Y when there are 2 equations. Both equations have an X and a Y so that means there are 4 X and Y answers.

Mathematics
1 answer:
AveGali [126]3 years ago
6 0

Answer:

x = 1

y = -4

Step-by-step explanation:

We will multiply the 2nd equation by 4 first:

4 * [-x + y = -5]

-4x + 4y = -20

Now we will add this equation with 1st equation given. Shown below:

4x + 3y = -8

-4x + 4y = -20

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

7y = -28

Now, we can solve for y easily:

7y = -28

y = -28/7

y = -4

Now, we take this value of y and put it in 1st original equation and solve for x:

4x + 3y = -8

4x + 3(-4) = -8

4x - 12 = -8

4x = -8 + 12

4x = 4

x = 4/4

x = 1

So, this is the only solution to this problem ( 1 intersection point at x = 1 and y = -4)

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HELP QUICK PLEASE!
-Dominant- [34]

Answer:

Side-Side-Side (SSS)

Step-by-step explanation:

Side-Side-Side is a rule used to prove whether a given set of triangles are congruent. The SSS rule states that: If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.

4 0
3 years ago
g The tangent plane to z=f(x,y) at the point (1,2) is z=5x+2y−10. (a) Find fx(1,2) and fy(1,2). fx(1,2)= Number fy(1,2)= Number
murzikaleks [220]

Answer:

The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.

Approximation at points (1.1,1.9) is 0.7

Step-by-step explanation:

Given:

Tangent plane to  a surface z=5x+2y-10 as the function at point (1,2)

To find :

f(x,y) at (1,2)

partial derivatives of function w.r.t. (x and y) and value of that function at given points.

Solution:(refer the attachment also)

Now we know that

the equation of tangent plane at given points to the surface is given by,

f(x1,y1,z1) and z=f(x,y)

z-z1=Fx(x1,y1)*(x-x1)+Fy(x1,y1)*(y-y1)

here Fx(x1,y1) and Fy(x1,y1) are the partial derivatives of x and y.

now

taking partial derivative w.r.t. x we get

Fx(x1`,y1)=\frac{d}{dx} (5x+2y-10)

=5.

Then w.r.t y we get

Fy(x1,y1)=

\frac{d}{dy}(5x+2y-10)

=2.

The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.

Using the Linearization or linear approximation we get

L(x,y)=f(x1,y1)+Fx(x,y)*(x-x1)+Fy(x,y)(y-y1)

=-1+5(x-1)+2(y-2)

=5x+2y-10

Approximation at F(1.1,1.9)

=5(1.1)+2(1.9)-10

=5.5+3.8-10

=0.7

Approximation at points (1.1,1.9) is 0.7

6 0
3 years ago
A store sold 50 copies of a magazine for $150. Each copy of the magazine costs the same. Which equation and set of ordered pairs
snow_tiger [21]
It is 3 copies per magazine
4 0
3 years ago
Find the missing section. Answers must be in fraction form.
frozen [14]

Answer:

\frac{1}{10}

Step-by-step explanation:

We have a circle that is split in three sections, two of which we know and we are asked to find the third missing section.

For the circle, we know that 4/5 and 1/10 is fit. Now we need the last one, to solve, we need to get the same denominator and see how much is missing.

Since 1/10 is our highest denominator, let's change 4/5 to have 10 as a denominator. Which would be through multiplying 5 to get 10.

What times 5 equals 10?

2

Now multiply both numerator and denominator by 2 to get our portion.

\frac{4*2}{5*2}

\frac{8}{10}

Now we have the same denominator, let's add our two fractions and see how much we have left.

8/10 + 1/10

9/10

We have 1/10 missing, therefore 1/10 is the answer.

3 0
2 years ago
Which equation is equivalent to √x^2+81 = x+10
jeka57 [31]

Keywords:

<em>equation, operations, equivalent, binomial, square root </em>

For this case we have an equation in which we must apply operations to rewrite it in an equivalent way. We must start by raising both sides of the equation to the square. Thus, we eliminate the square root of the left side of equality and finally solve the binomial of the right side of equality.

So we have:

\sqrt {x ^ 2 + 81} = x + 10\\(\sqrt {x ^ 2 + 81}) ^ 2 = (x + 10) ^ 2\\x ^ 2 + 81 = (x + 10) ^ 2

By definition:(a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2

x ^ 2 + 81 = x ^ 2 + 2 (x) (10) + (10) ^ 2\\x ^ 2 + 81 = x ^ 2 + 20x +100

Thus, x ^ 2 + 81 = x ^ 2 + 20x +100 is equivalent to \sqrt {x ^ 2 + 81} = x + 10

Answer:

x ^ 2 + 81 = x ^ 2 + 20x +100

Option D

4 0
3 years ago
Read 2 more answers
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