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3 years ago

Find the slope please

Mathematics

1 answer:

Katarina [22]3 years ago

7 0

Answer:

slope is $-\frac{3}{4}$

Step-by-step explanation:

1,-1 -3,2 are two points on the graph

the equation to find slope is $\frac{y_{1}- y_{2} }{x_{1}- x_{2} }$

so $\frac{2--1}{-3-1} = -\frac{3}{4}$

or you can count how many places it changed in the equation $\frac{rise}{run}$

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Write the decomposition that helps you, and then round to the given place value. Draw number lines to explain your thinking. Cir

egoroff_w [7]

Further explanation

There are general rules used in rounding numbers

1. Determine the rounded digit

2. Pay attention to the digits afterward (right). If the digit shows the numbers 0, 1, 2, 3 or 4, then the digit to be rounded is fixed, and the digit on the right is 0 all

3. If the digit on the right shows the numbers 5, 6, 7, 8 or 9, then the digit to be rounded is added to one, and the digit on the right side is made all 0

In rounding the nearest hundreds number, then we see the digit in the number tens (right of the digit hundreds), if it is worth 0, 1, 2, 3 or 4, then it remains, but if it shows numbers 5, 6, 7, 8 or 9, then the digit to be rounded is added one, and the right-hand side is 0

Calculated from the decimal point, the place on the right represents tents, hundredths, thousandths, etc.

a. 32.697

Decomposition

ten : 3

one : 2

tenth : 6

hundredth : 9

thousandths : 7

For the tenth number is 6, the value on the right is 9 then rounded become 32.7

For the hundredth number is 9, the value on the right is 7 then rounded become 32.70

for the one number is 2 , the value on the right is 6 then rounded become 33

b. 141.999

Decomposition

hundred : 1

ten : 4

one : 1

tenth : 9

hundredth : 9

thousandths : 9

a. For the tenth number is 9, the value on the right is 9 then rounded become 142.0

For the hundredth number is 9, the value on the right is 9 then rounded become 142.00

For the ten number is 4, the value on the right is 1 then rounded become

For the ten number is 1, the value on the right is 4 then rounded become

142

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3 years ago

Sanji scored 125 points in the first round of a video game and 263 points in the second round. His total score after the third r

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Sanji scored 44 points more in the third round than in the first round.

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2 years ago

Write a sine or cosine function for the data set

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Answer:

though this is difficult i know you would get it. Anyhow im real tired... will solve later... gl

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3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

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2 years ago

Anyone know the answer???

Triss [41]

The rate of change is 4. A linear model.

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3 years ago