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3 years ago

Which ordered pairs are solutions to the equation 2x + 10y = 12?

Mathematics

2 answers:

White raven [17]3 years ago

5 0

A pair could be (4,2/5)

vovangra [49]3 years ago

4 0

Answer:

(4,2/5) (−5,11/5)

Step-by-step explanation:

To determine which pairs are solutions, substitute the coordinates into the left side of the equation and if equal to the right side then they are solutions.

(7, - 2)

2(7) + 10(- 2) = 14 - 20 = - 6 ≠ 12 → (7, - 2) is not a solution

(- 5, - 4)

2(- 5) + 10(- 4) = - 10 - 40 = - 50 ≠ 12 → (- 5, - 4) is not a solution

(4, )

2(4) + 10() = 8 + 4 = 12 → hence solution

(- 5, )

2(- 5) + 10( ) = - 10 + 22 = 12 → hence solution

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Find the equation of a line with slope 1/2 through the point (10,12)

mihalych1998 [28]

Answer:

$x-2y+14=0$

Step-by-step explanation:

$Slope \:( m) =\frac{1}{2}$

$(x_1 \:y_1) =(10, \:12)$

Equation of line in point slope form is given as:

$y-y_1 =m(x-x_1)$

Plugging the values of m and $(x_1 \:y_1)$ , we find:

$y-12=\frac{1}{2} (x-10)$

$2y-24=x-10$

$0=x-10-2y+24$

$x-2y+14=0$

This is the required equation of line.

5 0

3 years ago

Pls help. Take a pic of your work on paper thanks.

artcher [175]

Answers:

1.) -x - 5

2.) 11x^2 + 5x

3.) 13x + 2y +3y^2

4.) -12x - 2

5.) -24

Sorry, it won’t somehow let me attach a picture of my work on paper :(

But hopefully this helps! :)

6 0

3 years ago

Read 2 more answers

What are the limits of integration if the summation the limit as n goes to infinity of the summation from k equals 1 to n of the

Fofino [41]

Answer:

$\int_{2}^{9}x^2 dx$ so the limits are 2 and 9

Step-by-step explanation:

We want to express $\lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2$ as a integral. To do this, we have to identify $\sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2$ as a Riemann Sum that approximates the integral. (taking the limit makes the approximation equal to the value of the integral)

In general, to find a Riemann sum that approximates the integral of a function f over an interval [a,b] we can the interval in n subintervals of equal length and approximate the area (integral) with rectangles in each subinterval and them sum the areas. This is equal to

$\sum_{k=1}^n f(y_k) \frac{b-a}{n}$ , where $y_k\in[a+(k-1)\frac{b-a}{n},a+k\frac{b-a}{n}]$ is a selected point of the subinterval.

In particular, if we select the ending point of each subinterval as the $y_k$ , the Riemann sum is:

$\sum_{k=1}^n f(a+k\frac{b-a}{n}) \frac{b-a}{n}$ .

Now, let's identify this in $\sum_{k=1}^n\frac{1}{7n}(2+\frac{7k}{n})^2$ .

The integrand is x² so this is our function f. When k=n, the summand should be $\frac{b-a}{n}f(b)=\frac{b-a}{n}b^2$ because the last selected point is b. The last summand is $\frac{7}{n}(9)^2$ thus b=9 and b-a=7, then 9-a=7 which implies that a=2.