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3 years ago

12

The record high January temperature in Austin,Texas ,90f .The record low January temperature -2f . Find the difference between t

he high and low temperatures

2 answers:

Fudgin [204]3 years ago

7 0

Answer:

The distance between 90 and -2 is 92

stich3 [128]3 years ago

4 0

You might think that you would subtract but you are really just trying to find the space between them. Think of a thermometer and draw one if it helps. Mark 90 °F about where it would be and then mark -2°F remembering to put it below 0. Now find the distance between them by adding. The distance between 90 and -2 is 92.

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Differentiate the following by using "limit"

Nat2105 [25]

Answer:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ 2\sqrt{x } }$

Step-by-step explanation:

we want to differentiate the following by using limit:

$\displaystyle \frac{d}{dx} \sqrt{x}$

derivative definition by limit given by

$\rm \displaystyle \frac{d}{dx} = \lim _{\Delta x \to 0} \left( \frac{f(x + \Delta x) - f(x)}{ \Delta x} \right)$

given that,

f(x)=√x

so,

f(x+∆x)=√(x+∆x)

thus substitute:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \sqrt{x + \Delta x}- \sqrt{x} }{ \Delta x} \right)$

multiply both the numerator and denominator by the conjugate of the numerator:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \sqrt{x + \Delta x}- \sqrt{x} }{ \Delta x} \times \frac{ \sqrt{x + \Delta x} + \sqrt{x} }{\sqrt{x + \Delta x} + \sqrt{x}} \right)$

simplify which yields:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ (\sqrt{x + \Delta x}) ^{2} - x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right)$

simplify square:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{ \Delta x \to 0} \left( \frac{ x + \Delta x - x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right)$

collect like terms:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \Delta x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right)$

reduce fraction:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ 1 }{ (\sqrt{x + \Delta x} + \sqrt{x})} \right)$

get rid of ∆x as we are approaching its to 0

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ \sqrt{x } + \sqrt{x}}$

simplify addition:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ 2\sqrt{x } }$

and we are done!

7 0

3 years ago

Can you help me please(:

kherson [118]

6(10)=6 times 10 which is 60

4 0

4 years ago

Read 2 more answers

Me and my daughter are having a disagreement on this question, I would appreciate the help.

elena-14-01-66 [18.8K]

Answer:

(4t - 8/5) - (3 - 4/3t) = 16/3t - 23/5

5(2t + 1) + (-7t + 28)= 3t+33

(-9/2t +3) + (7/4t + 33) = -11/4c+56

3 (3t-4) - (2t + 10)= 7t -22

Sooo Sorry that it took soooo long! Hope this helps!!!

Happy holidays!!

4 0

3 years ago

AnnZ [28]

Part a: The value of x is 33°

Part b: The measure of each base angle is 69°

Explanation:

It is given that the vertex angle of an isosceles triangle measures 42°.

The base angle in the triangle has a measure of $(2x+3)^{\circ}$

Part a: To determine the value of x

Since, in an isosceles triangle the base angles are always equal.

Thus, the three angles in an isosceles triangle add up to 180°

Hence, we have,

$42+2x+3+2x+3=180$

$42+4x+6=180$

$4x+48=180$

$4x=132$

$x=33^{\circ}$

Thus, the value of x is 33°

Part b: To find the measure of each base angle.

Base angle = $(2 x+3)^{\circ}$

Substituting $x=33^{\circ}$ , we get,

$Base \ angle = (2(33)+3)^{\circ}$

$=(66+3)^{\circ}$

$=69^{\circ}$

Thus, the measure of each base angle is 69°

6 0

4 years ago

What is the multiplication form six to the fourth power to the second power

igomit [66]

Answer:

1679616

Step-by-step explanation:

(6^4)^2

6*6*6*6*6*6= 1296

1296^2 = 1296*1296

= 1679616

hope this helps :)

5 0

3 years ago