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2 years ago

5

Antonio graphs these equations and finds that the lines intersect at a single point, (–5, 0.25). Equation A: Equation B: 4y – 3x

= 16 –x – 8y = 3 Which statement is true about the values x = –5 and y = 0.25?

2 answers:

Elza [17]2 years ago

6 0

Given equations : Equation A: 4y -3x =16 and

Equation B: -x-8y =3.

We are given a<em> point (-5,0.25) plugging it in first equation</em>

4(0.25) -3(-5) =16

1+15 =16.

16 =16 <em>: Satisfies first equation.</em>

Plugging (-5,0.25) it in second equation.

-(-5)-8(0.25) =3.

5 - 2 = 3.

3=3 <em>: Satisfies first equation.</em>

<h3>Therefore, they are the only values that make both equations true.</h3>

Firlakuza [10]2 years ago

5 0

Answer:

"they are the only values that make both equations true."

Step-by-step explanation:

still the correct answer on APEX as of 4/29/2019

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The area of a triangle can be represented by the expression 14x^5+63x^2. If the base is 7x^2, write an expression to represent i

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2 years ago

Della bought a tree seedling that was 2 1/4 feet tall. During the first year, it grew 1 1/6 Feet. After two years, it was 5 feet

Margarita [4]

Answer:

Step-by-step explanation:

Alright, lets get started.

Della bought a tree seeding that was $2\frac{1}{4}$ feet tall means $\frac{9}{4}$ feet.

First year it grew $1\frac{1}{6}$ feet means it grew $\frac{7}{6}$ feet.

It means after 1 year, the height of tree will be = $\frac{9}{4}+\frac{7}{6}=\frac{41}{12}$

After 2 years, the height of tree is = 5 feet

So, it grew in second year = $5-\frac{41}{12}$

So, it grew in second year = $\frac{60-41}{12}=\frac{19}{12}$

So, it grew in second year= $1\frac{7}{12}$ feet. : Answer

Hope it will help :)

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2 years ago

Charlotte’s weekly paycheck is based on the number of hours worked during the week and on the weekend. If she works 13 hours dur

Gelneren [198K]

I am setting the week hourly rate to x, and the weekend to y. Here is how the equation is set up:

13x + 14y = $250.90
15x + 8y = $204.70

This is a system of equations, and we can solve it by multiplying the top equation by 4, and the bottom equation by -7. Now it equals:

52x + 56y = $1003.60
-105x - 56y = -$1432.90

Now we add these two equations together to get:

-53x = -$429.30 --> 53x = $429.30 --> (divide both sides by 53) x = 8.10. This is how much she makes per hour on a week day.

Now we can plug in our answer for x to find y. I am going to use the first equation, but you could use either.

$105.30 + 14y = $250.90. Subtract $105.30 from both sides --> 14y = $145.60 divide by 14 --> y = $10.40

Now we know that she makes $8.10 per hour on the week days, and $10.40 per hour on the weekends. Subtracting 8.1 from 10.4, we figure out that she makes $2.30 more per hour on the weekends than week days.

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2 years ago

Read 2 more answers

A fast-food restaurant has a cost of production C(x)=12x+114 and a revenue function R(x)=6x. When does the company start to turn

LekaFEV [45]

Step-by-step explanation:

Given data

C(x)=12x+114

R(x)=6x.

The company will start to run on profit just when revenue equals the production cost, that is

C(x)=R(x)

equating the two expressions we have

12x+114=6x

solving for x we have

12x-6x+114=0

6x+114=0

6x=-114

x=-114/6

x=-19

Hence the company will start to run on proft when they start to produce above 19 items

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2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

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2 years ago