All categories

3 years ago

Solve for x and y: y=2+7x −2x+6y=172 Select one:

Mathematics

1 answer:

Ymorist [56]3 years ago

5 0

It is pretty easy to eliminate the wrong answers by trying the answers in the equations.

a. 7 times 30 is way more than 4

c. 6 times 226 is way more than 172, even after subtracting 64

d. 7 times -74 is way less than 4

The appropriate choice is

... b. x=4, y=30

You can check it with your graphing calculator.

_____

Since you have an expression for y, you can substitute that into the second equation:

... -2x + 6(2+7x) = 172 . . . . substitute for y

... 40x = 160 . . . . . . . . . . . . eliminate parentheses, subtract 12, collect terms

... x = 4 . . . . . . this is sufficient to help you locate the correct answer

You might be interested in

A square with a side of 74 units.

lidiya [134]

Answer: In the text

Step-by-step explanation:

The length of a rectangle is twice the width. The area of the rectangle is 74 square units. Notice that you can divide the rectangle into two squares with equal area. How can you estimate the side length of each square? Estimate the length and width of the rectangle.

3 0

2 years ago

The diagram shows a logo made from three circles

Nana76 [90]

Answer:

3cm+2cm+5cm=10cm

%=100

10*10=100

2cm*10=20%

Step-by-step explanation:

the shaded area is width of 2 2/10 or 20/100

8 0

4 years ago

Read 2 more answers

AP Calculus redo! I know the answer but can't figure out exactly how to get there. Thank you! I want to work through the steps.

Monica [59]

You're approximating

$\displaystyle\int_1^5 x^2\,\mathrm dx$

with a Riemann sum, which comes in the form

$\displaystyle\int_a^b f(x)\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x_i$

where $x_i$ are sample points chosen according to some decided-upon rule, and $\Delta x_i$ is the distance between adjacent sample points in the interval.

The simplest way of approximating the definite integral is by partitioning the interval into equally-spaced subintervals, in which case $\Delta x=\dfrac{b-a}n$ , and since $[a,b]=[1,5]$ , we have

$\Delta x=\dfrac{5-1}n=\dfrac4n$

Using the right-endpoint method, we approximate the area under $f(x)$ with rectangles whose heights are determined by their right endpoints. These endpoints are chosen by successively adding the subinterval length to the starting point of the interval of integration.

So if we had $n=4$ subintervals, we'd split up the interval of integration as

$[1,5]=[1,2]\cup[2,3]\cup[3,4]\cup[4,5]$

Note that the right endpoints follow a precise pattern of

$2=1+\dfrac44$
$3=1+\dfrac84$
$4=1+\dfrac{12}4$
$5=1+\dfrac{16}4$

The height of each rectangle is then given by the values above getting squared (since $f(x)=x^2$ ). So continuing with the example of $n=4$ , the Riemann sum would be

$\displaystyle\sum_{i=1}^4\left(1+\dfrac{4i}4\right)^2\dfrac44$

For $n=5$ ,

$\displaystyle\sum_{i=1}^5\left(1+\dfrac{4i}5\right)^2\dfrac45$

and so on, so that the definite integral is given exactly by the infinite sum

$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\left(1+\dfrac{4i}n\right)^2\dfrac4n$