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2 years ago

Select the linear equation represented in the graph.

Mathematics

1 answer:

AnnZ [28]2 years ago

4 0

To find the linear equation, use the slope formula y = mx+b, where b is the value of the y-intercept and m is rise/run.

The line rises 1 unit and runs (left to right) -5 units. (you could also say the line runs -1 units and runs 5 units, they will both give the same answer). Rise/run, 1/-5 = -1/5.

The y-intercept is the y value where the line touches the y-axis. In this case, it is +1.

Plug the information into the equation. m = -1/5 and y = 1. y = -1/5x + 1

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dnw snow storm Francisco had 17 inches of snow on his lawn.the temperature then increased in the snow begin to melt at a consist

stich3 [128]

Answer:

Step-by-step explanation:

(equation: 17-2.5=2)

17-2.5=2

-2.5

17=2

17 divided by 2

8.5 is the

he would have 8.5 inches of snow on his lawn.

4 0

2 years ago

A palindrome is a number that reads the same forward and backward. For example, 2442 and 111 are palindromes. If 5-digit palindr

Verdich [7]

Answer:

Step-by-step explanation:

Let the 5 digit palindrome formed from 1,2,3,4,5 be represented as XYZXY

Possible outcomes of X are 1,2,3 which give 3 possible outcomes

Possible outcomes of Y are 1,2,3 which also gives 3 possible outcomes

Possible outcomes of Z are 1,2,3 which will give 3 possible outcomes.

The total possible outcome = 3*3*3

= 27

6 0

2 years ago

A square has side lengths of 3/4 in. What is its area?

Leno4ka [110]

Hey there! :)

Answer:

A = 9/16 in².

Step-by-step explanation:

Use the formula A = s² to solve for the area where s = 3/4 in:

A = (3/4)²

A = 9/16 in².

7 0

3 years ago

Read 2 more answers

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago

A set of ordered pairs is called ordered pairs is called

Amiraneli [1.4K]

A set of ordered pairs is called a relation. The set of all first components of the ordered pairs of a relation is the domain of the relation, and the set of all second components of the ordered pairs is the range of the relation. :)

4 0

3 years ago