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3 years ago

Amy says that the 9 in the number 900.876 is 1/1,000 the size of the 9 in 6.91. Is she correct? How do you know?

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

8 0

Answer:

Step-by-step explanation:

The 9 in 900.876 is 900 and the 9 in 6.91 is .9 and if you look at the place value, you'll see that 900 is larger than .9

Phoenix [80]3 years ago

4 0

Answer: The 9 in 900.876 is 900 and the 9 in 6.91 is .9 and if you look at the place value, you'll see that 900 is larger than 9. so no

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Which equation describes the line that passes through the points (-4, 19) and (2, -1)

dangina [55]

Answer:

y=-5/2x+4

Step-by-step explanation:

find the slope by using y2-y1/x2-x1

-1-19/2-(-4)

simplify

-20/8

simplify

-5/2

use slope-intercept form, y=mx+b

since we know the slope, find b

plug in one of the ordered pairs into the equation

-1=(-5/2)(2)+b

simplify

-1= -10/2+b

simplify

-1=-5+b

add 5 to both sides

b=4

plug b into y=-5/2x+b

y=-5/2x+4

8 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

Devin ran 6 over 10 miles on Thursday.He ran 1.5 miles on Saturday. How many total miles didi he run

Diano4ka-milaya [45]

Answer:

17.5 miles

Step-by-step explanation:

6+10+1.5= 17.5

8 0

3 years ago

What is 2 times 3 plus 57 - 100

kirill [66]

Answer:

-37

Step-by-step explanation:

2 x 3 + 57 -100

we do something called pemdas

We start with multiplying first

2 x 3 = 6

Then we do addition

6+57 = 63

then we subtract

63- 100 = -37

7 0

3 years ago

Read 2 more answers

I need help with this Factoring Quadratics question.Can someone help me?

antoniya [11.8K]

If you mean "factor over the rational numbers", then this cannot be factored.

Here's why:

The given expression is in the form ax^2+bx+c. We have

a = 3

b = 19

c = 15

Computing the discriminant gives us

d = b^2 - 4ac

d = 19^2 - 4*3*15

d = 181

Note how this discriminant d value is not a perfect square

This directly leads to the original expression not factorable

We can say that the quadratic is prime

If you were to use the quadratic formula, then you should find that the equation 3x^2+19x+15 = 0 leads to two different roots such that each root is not a rational number. This is another path to show that the original quadratic cannot be factored over the rational numbers.

7 0

3 years ago