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3 years ago

Graph the function. y=2-x

Mathematics

1 answer:

oee [108]3 years ago

4 0

Its the 3rd from the left good luck

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Callie spends $36 on fabric. Each yard costs $3. How much fabric does Callie buy?

Evgesh-ka [11]

So you divide 36 by 3 and get 12. Therefore, Callie bought 12 yards of fabric

7 0

4 years ago

Read 2 more answers

A (0, 2) and B (6,6) are points on the straight line ABCD.

elena-14-01-66 [18.8K]

Answer:

(18, 14)

Step-by-step explanation:

We know that C and D lie on the line AB and BC = CD = AB. Then we need to use the distance formula and equation of the line AB to find the other two coordinates.

The distance formula states that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ , the distance is denoted by: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Let's find the distance between A and B:

d = $\sqrt{(6-0)^2+(6-2)^2}=\sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52} =2\sqrt{13}$

Now say the coordinates of D are (a, b). Then the distance between D and B will be twice of 2√13, which is 4√13:

4√13 = $\sqrt{(6-a)^2+(6-b)^2}$

Square both sides:

208 = (6 - a)² + (6 - b)²

Let's also find the equation of the line AB. The y-intercept we know is 2, so in y = mx + b, b = 2. The slope is (6 - 2) / (6 - 0) = 4/6 = 2/3. So the equation of the line is: y = (2/3)x + 2. Since (a, b) lines on this line, we can put in a for x and b for y: b = (2/3)a + 2. Substitute this expression in for b in the previous equation:

208 = (6 - a)² + (6 - b)²

208 = (6 - a)² + (6 - (2/3a + 2))² = (6 - a)² + (-2/3a + 4)²

208 = a² - 12a + 36 + 4/9a² - 16/3a + 16 = 13/9a² - 52/3a + 52

0 = 13/9a² - 52/3a - 156

13a² - 156a - 1404 = 0

a² - 12a - 108 = 0

(a + 6)(a - 18) = 0

a = -6 or a = 18

We know a can't be negative so a = 18. Plug this back in to find b:

b = 2/3a + 2 = (2/3) * 18 + 2 = 12 + 2 = 14

So point D has coordinates (18, 14).

8 0

4 years ago

Read 2 more answers

28 points and brainliest hurry!!

Sophie [7]

A= -2,4

B= -2,4

C= 2,7

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%

garri49 [273]

Answer:

<h3>METHOD I:</h3>

(by using the first principle of differentiation)

We have the "Limit definition of Derivatives":

$\boxed{\mathsf{f'(x)= \lim_{h \to 0} \{\frac{f(x+h)-f(x)}{h} \} ....(i)}}$

Here, f(x) = sec x, f(x+h) = sec (x+h)

Substituting these in eqn. (i)

$\implies \mathsf{f'(x)= \lim_{h \to 0} \{\frac{sec(x+h)-sec(x)}{h} \} }$

sec x can be written as 1/ cos(x)

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{1}{cos(x+h)} -\frac{1}{cos(x)} \} }$

Taking LCM

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{cos(x)-cos(x+h)}{cos(x)cos(x+h)} \} }$

By Cosines sum to product formula, i.e.,

$\boxed{\mathsf{cos\:A-cos\:B=-2sin(\frac{A+B}{2} )sin(\frac{A-B}{2} )}}$

=> cos(x) - cos(x+h) = -2sin{(x+x+h)/2}sin{(x-x-h)/2}

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{2sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{sin(\frac{h}{2} )}{h} }$

I shifted a 2 from the first limit to the second limit, since the limits ar ein multiplication this transmission doesn't affect the result

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{2sin(\frac{h}{2} )}{h} }$

2/ h can also be written as 1/(h/ 2)

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{1\times sin(\frac{h}{2} )}{\frac{h}{2} } }$

We have limₓ→₀ (sin x) / x = 1.

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: 1 }$

h→0 means h/ 2→0

Substituting 0 for h and h/ 2

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+0)}{cos(x+0)cos(x)} }$

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)cos(x)} }$

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)}\times \frac{1}{cos x} }$

sin x/ cos x is tan x whereas 1/ cos (x) is sec (x)

$\implies \mathsf{f'(x)= tan(x)\times sec(x) }$

Hence, we got

$\underline{\mathsf{\overline{\frac{d}{dx} (sec(x))=sec(x)tan(x)}}}$

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

<h3>METHOD II:</h3>

(by using other standard derivatives)

$\boxed{ \mathsf{ \frac{d}{dx} ( \sec \: x) = \sec x \tan x }}$

sec x can also be written as (cos x)⁻¹

We have a standard derivative for variables in x raised to an exponent:

$\boxed{ \mathsf{ \frac{d}{dx}(x)^{n} = n(x)^{n - 1} }}$

Therefore,

$\mathsf{ \frac{d}{dx}( \cos x)^{ - 1} = - 1( \cos \: x) ^{( - 1 - 1} } \\ \implies \mathsf{\ - 1( \cos \: x) ^{- 2 }}$

Any base with negative exponent is equal to its reciprocal with same positive exponent

$\implies \: \mathsf{ - \frac{1}{ (\cos x) {}^{2} } }$

The process of differentiating doesn't just end here. It follows chain mechanism, I.e.,

while calculating the derivative of a function that itself contains a function, the derivatives of all the inner functions are multiplied to that of the exterior to get to the final result.