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2 years ago

6

What is the decimal expansion of the number 28 over 4

1 answer:

Gre4nikov [31]2 years ago

7 0

28/4 = 7
there is no decimal expansion

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HELP ME PLEASE, i will give brainliest

max2010maxim [7]

Based on the right angle triangle and the given parameters, cos K = 5/13

<h3>Trigonometric ratios</h3>

sin = opposite / hypotenuse

cos = adjacent / hypotenuse

Tan = opposite / adjacent

Given parameters:

Hypotenuse = 13
Adjacent = 5
Opposite = 12

Therefore,

cos K = adjacent / hypotenuse

cos K = 5/13

Learn more about right triangle:

brainly.com/question/2217700

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2 years ago

in a survey, 250 adults and children were asked whether they know how to swim. the surgery data are shown in the relative freque

Nadusha1986 [10]

Answer: D

Step-by-step explanation:

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2 years ago

Solve for x. Show your work.

hram777 [196]

5(5+x) = 6(6+4)
25+5x = 60
5x = 35
x = 7

-----------------
3(3+5) = 4(4+x)
24 = 16 + 4x
4x = 24-16
4x = 8
x = 2

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3 years ago

Read 2 more answers

HELPPP ILL GIVE EXTRA POINTS QUESTION ABOVE

MatroZZZ [7]

Answer:

The answer is the top option choice "both i and ii."

Step-by-step explanation:

Hopes this helps.

8 0

2 years ago

Pls, can you help me? thx.<br><img src="https://tex.z-dn.net/?f=%20%5Ccos%28x%20%2B%20%5Cfrac%7B%5Cpi%7D%7B3%7D%29%20%5Cgeqslant

Veseljchak [2.6K]

For $x$ between $-\pi$ and $\pi$ , we have

$\cos x=\frac{\sqrt2}2=\frac1{\sqrt2}$ when $x=\pm\frac\pi4$ ;
$\cos x=1$ for $x=0$ ; and
$\cos x=0$ for $x=\pm\frac\pi2$

$\cos x$ is continuous over its domain, so the intermediate value theorem tells us that

$\cos x\ge\frac{\sqrt2}2$

is true for $-\frac\pi4\le x\le\frac\pi4$ .

For all $x$ , we take into account that $\cos x$ is $2\pi$ -periodic, so the above inequality can be expanded to

$-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4$

where $n$ is any integer. Equivalently,

$-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi$

To get the corresponding solution set for

$\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2$

simply replace $x$ with $x+\frac\pi3$ :

$-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi$

$\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}$

7 0

3 years ago