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3 years ago

What is 1x1? please help me lol

Mathematics

2 answers:

Ksju [112]3 years ago

7 0

Answer:

Step-by-step explanation:

1 times anything is itself

you're welcome

Andreyy893 years ago

6 0

Answer:

Step-by-step explanation:

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Is this true or False? (0,8) is an x-intercept. If False, provide a correction.

pav-90 [236]

Answer:

True

Step-by-step explanation:

This is true because it crosses the x-axis. (x,y)

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2 years ago

create three different drawings showing a number of rectangles and circle on which the ratio of rectangle to circles is 3;1

andre [41]

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3 years ago

Will has 40 state quarters if 3/5 of his quarters are dated 2005 what is the value of the quarters from 2005?

san4es73 [151]

Answer:

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3 0

3 years ago

Pls help me with trig :)

trapecia [35]

Answer:

Step-by-step explanation:

let ∠GEF=x

then ∠GFE=90-x

$\frac{GH}{EH}=tan x\\GH=EH~ tan x\\or~ GH=a ~tan x=6 tan x\\\frac{GH}{FH}=tan (90-x)=cot x\\GH=FH*cot x\\GH=b*cot x=27 cot x\\6 tan x=27 cotx\\\frac{tan x}{cot x}=\frac{27}{6}\\tan ^2 x=\frac{27}{6}=\frac{9}{2}\\sec^2x-tan^2x=1\\sec^2 x=1+tan ^2x=1+\frac{9}{2}=\frac{11}{2}\\\frac{EG}{EH}=sec x\\ EG=EH sec x=a sec x= 6 sec x=6\sqrt{\frac{11}{2} } =3\sqrt{22} \approx 14.07$

4 0

3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

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