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3 years ago

Slope intersect form

Mathematics

1 answer:

pochemuha3 years ago

6 0

Well... you need to find your slope first for starters. In order to find you slope it would be y2-y1/x2-x1... rise/run. So that would make your slope 6/3 and you can simplify that to 2/1 because divide it by 3. In order to find your y-intercept, you need to have (x) at 0 and wherever (y) is on that line, it would be your y-intercept.

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the sum of four consecutive integers is 2174. what are the integers? (continuation) the smallest of four consecutive integers is

timofeeve [1]

The smallest of the four numbers is x.
Since the four numbers are consecutive, this means that each number is one more than the previous.
Therefore, the four numbers are:
x
x + 1
x+1 + 1 = x+2
x+2 + 1 = x+3

Now, we are given that the sum (s) of the four numbers is 2174
This means that:
s = x+x+1+x+2+x+3
s = 4x + 6

We are given that s = 2174. Substitute with s in the above equation and solve for x as follows:
2174 = 4x + 6
4x = 2174 - 6
4x = 2168
x = 2168 / 4
x = 542

Based on the above calculations, the four numbers are:
542, 543, 544 and 555

7 0

2 years ago

Read 2 more answers

The fuel cost per hour for running a ship is approximately one half the cube of the speed (measured in knots) plus additional fi

andrezito [222]

Answer:

6 knots

Step-by-step explanation:

Let the speed be v knots

then time taken to cover 500 M = 500 / v hrs

fuel consumption /hr = 216 + 0.5v^3

let F be the fuel consumption for trip

= [500/v][216 + 0.5v^3]

= 500[216/v + 0.5v^2]

dF/dv = 500[ - 216/v^2 + v]

d^2F/d^2v = 500[432/v^3 + 1] , i.e. +ve

so setting dF/dv will give a minima

500[ -216/v^2 + v] = 0

or v = 216/v^2

or v^3 = 216

solving, we get v = [216]^(1/3) = 6 knots

7 0

3 years ago

Can you help me please!!!!

pentagon [3]

If you add the 3/4s with the 2 1/4 hour breaks it would be 1 hour and 1 fourth.
Then you take 10-1/2 hours and subtract it by 1 hour and 1 fourth.
So you would get 9-1/4 hours

5 0

3 years ago

Read 2 more answers

What is 15ft by 18ft times 2in depth converted into yards

NeTakaya

Answer:

5 yd³

Step-by-step explanation:

2 in = 1/6 ft

15 ft × 18 ft × 1/6 ft = 45 ft³

1 ft³ = 9 yd³

45 ft³/9 = 5 yd³

4 0

3 years ago

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago