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2 years ago

Which expressions have a value greater than 8?

Mathematics

2 answers:

REY [17]2 years ago

6 0

Answer:

A lot of expressions have a greater value that 8.

Step-by-step explanation:

strojnjashka [21]2 years ago

5 0

Answer:

So 8 more than 'x' is: x + 8.

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Terry, Kevin and barry together have 215 marbles. Terry has 20 marbles more than Barry and kevin has 10 marbles more than Terry.

Zarrin [17]

Terry, Kevin and Barry have a total of 215 marbles.=> Terry has 20 marbles more than Barry = x + 20=> Kevin has 10 marbles more than Terry = (x + 20) + 10=> Barry = xSolutions:=> x + x + 20 + x + 20 + 10 = 215=> 3x + 50 = 215=> 3x = 215 - 50=> 3x = 165=> 3x / 3 = 165 / 3=> x = 55Barry have = 55Terry = 55 + 20 = 75<span>Kevin => 75 + 10 = 85</span>

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3 years ago

I WILL AWARD BRAINLIST PLS SOLVE

Ivanshal [37]

Answer:

Ah yes rsm.

Anyways, AB is parallel to DE so 72= angle BDE. So BDE+2x+2x=180

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2 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

2 years ago

When you raise (almost) anything to the power zero, you get 1.

larisa86 [58]

Answer:

z^0= 1

(2v+2)^0 = 1

3^0 = 1

0^0= 0

Step-by-step explanation:

When you raise (almost) anything to the power zero, you get 1 and when you raise 0 to any number except 0 is 0

6 0

1 year ago

Martha is carrying her biology, science, and math textbooks. Her books weigh a total of 9 pounds. Her biology textbook weighs 3

Nata [24]

1 pound = 16 oz
Her books weigh 9 pounds or (9)(16)= 144 oz
Math Textbook= 40 oz
Biology textbook= 3 pounds or (3)(16)= 48 oz

Total weight of books - math textbook - biology textbook = science textbook
144-40-48 = 56oz
science textbook = 56oz

5 0

3 years ago