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Oksanka [162]
3 years ago
5

What expression represents a number one fourth as great as 10-2

Mathematics
2 answers:
Iteru [2.4K]3 years ago
7 0

Answer:

The expression represents

a\geq\frac{1}{4}\times (10^{-2}) and solution is   a \geq 0.0025  

Step-by-step explanation:

Given : Expression represents a number one fourth as great as 10^{-2}

To find : What is the expression?

Solution :

Let the number be 'a'.

According to question,

A number one fourth as great as 10^{-2}

i.e.  a\geq \frac{1}{4}\times (10^{-2})

Now, we solve the expression

a \geq \frac{1}{4}\times\frac{1}{100}

a \geq\frac{1}{400}

a \geq 0.0025

Therefore, The expression represents

a\geq\frac{1}{4}\times (10^{-2}) and solution is   a \geq 0.0025

Bad White [126]3 years ago
6 0

Answer:

8+1/4

Step-by-step explanation:

8+1/4 because 10-2 is 8.

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Answer:

x = 4 , y = ⅕

Step-by-step explanation:

2x + 5y = 9 (1)

3x - 5y = 11 (2)

Add (1) and (2)

5x = 20

x = 4

Using (1), find y

2(4) + 5y = 9

5y = 1

y = ⅕

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3 years ago
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Find the exact value of cos(a+b) if cos a=-1/3 and cos b=-1/4 if the terminal side if a lies in quadrant 3 and the terminal side
maria [59]

Answer:

cos(a + b) = \frac{1}{12}(1-2\sqrt{30})

Step-by-step explanation:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]

cos(a) = -\frac{1}{3}

cos(b) = -\frac{1}{4}

Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.

sin(a) = -\sqrt{1-(-\frac{1}{3})^2} [Since, sin(a) = \sqrt{(1-\text{cos}^2a)}]

         = -\sqrt{\frac{8}{9}}

         = -\frac{2\sqrt{2}}{3}

Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be  negative.

sin(b) = -\sqrt{1-(-\frac{1}{4})^2}

         = -\sqrt{\frac{15}{16}}

         = -\frac{\sqrt{15}}{4}

By substituting these values in the identity,

cos(a + b) = (-\frac{1}{3})(-\frac{1}{4})-(-\frac{2\sqrt{2}}{3})(-\frac{\sqrt{15}}{4})

                = \frac{1}{12}-\frac{\sqrt{120}}{12}

                = \frac{1}{12}(1-\sqrt{120})

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Therefore, cos(a + b) = \frac{1}{12}(1-2\sqrt{30})

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3 years ago
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andrezito [222]

Answer:

a) \left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325

b) The lowest point of y=\cos \left(x\right), 0\leq x\leq 2\pi is when x = \pi

Step-by-step explanation:

a) To simplify the expression \left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right) you must:

Apply Difference of Two Squares Formula: \left(a+b\right)\left(a-b\right)=a^2-b^2

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\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)

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Apply the Pythagorean Identity 1+\tan ^2\left(x\right)=\sec ^2\left(x\right)

From the Pythagorean Identity, we know that 1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)

Therefore,

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b) According with the below graph, the lowest point of y=\cos \left(x\right), 0\leq x\leq 2\pi is when x = \pi

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See conversation in the attachment.

65 = 1000001₂

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