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andrey2020 [161]
2 years ago
14

How do I find slope intercept for a line that passes through (4,-4) with a slope of -1

Mathematics
2 answers:
yarga [219]2 years ago
3 0

Answer:

y = -x

Step-by-step explanation:

Use the equation:

y = mx + b

where m is the slope and b is the y-intercept.

In this case, there is no y intercept, as we can see by manipulating the point-slope equation:

y + 4 = -1(x + 4)

y = -x + 0

So, the equation for this line is y = -x.

skad [1K]2 years ago
3 0
Y=-1x
or
y=-1x + 0

Considering you already have the slope & a point you end up having

Y = mx + b

-4 which is your Y so

-4=mx + b

M is your slope so

-4=-1(x) + b

Your x is 4 so

-4= -1(4) + b

B is your y intercept but because -4 = -1 times 4 your b is zero
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<u>30 gallons solution 4% and 10 gallons solution 8%</u>

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<h3>x•4%+y•8%=40•5% </h3>

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4(40-y)+8y=200

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In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc
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Answer:

Step-by-step explanation:

(a)

Consider the following:

A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°

Use sine rule,

\frac{b}{a}=\frac{\sinB}{\sin A}&#10;\\\\=\frac{\sin{\frac{\pi}{3}}&#10;}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}

Again consider,

\frac{b}{a}=\frac{\sin{B}}{\sin{A}}&#10;\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Thus, the angle B is function of A is, B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Now find \frac{dB}{dA}

Differentiate implicitly the function \sin{B}=\sqrt{\frac{3}{2}}\sin{A} with respect to A to get,

\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}

b)

When A=\frac{\pi}{4},B=\frac{\pi}{3}, the value of \frac{dB}{dA} is,

\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}

c)

In general, the linear approximation at x= a is,

f(x)=f'(x).(x-a)+f(a)

Here the function f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

At A=\frac{\pi}{4}

f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}

And,

f'(A)=\frac{dB}{dA}=\sqrt{3} from part b

Therefore, the linear approximation at A=\frac{\pi}{4} is,

f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}

d)

Use part (c), when A=46°, B is approximately,

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