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Helga [31]
3 years ago
10

X minus 32? i need answer ASAP

Mathematics
2 answers:
Harrizon [31]3 years ago
6 0

Answer:

X-32 :)

Explanation:

Put together an expression.

tangare [24]3 years ago
5 0

Answer:

x-32

Step-by-step explanation:

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What is the completely factored form of this polynomial?
Sholpan [36]

Answer:

A. (x + 2)²(x - 2)²

Step-by-step explanation:

x⁴ - 8x² + 16 =

(x² - 4)(x² - 4) =

(x - 2)(x + 2)(x - 2)(x + 2) =

(x - 2)²(x + 2)²

5 0
3 years ago
Greatest common factor of 33 and 66
Firdavs [7]

the answer is 11 hope this helps

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3 years ago
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What is the area of ΔABC? Round to the nearest tenth of a square unit.
kramer
<span><u><em>Answer:</em></u>
8.4 square units

<u><em>Explanation:</em></u>
<u>The area of the triangle can be calculated using two sides and an included angle as follows:</u>
area = 0.5 * 1st side * 2nd side * sin(angle included between them)
............> This is shown in the attached image

<u>In the given triangle, we have:</u>
the two sides 2*</span>√2<span> and 6 units
the angle included between them is 80</span>°<span>

Therefore, we can apply the above rule to get the area as follows:
area = 0.5 * 2</span>√2 <span>*6* sin(80</span>°<span>)
area = 8.35 which is approximately 8.4 square units

Hope this helps :)</span>

4 0
3 years ago
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Helppp solve the equation algebraically
katen-ka-za [31]

Answer:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form:

( 3 , − 1 )

Equation Form:

x = 3 ,  y = − 1

Step-by-step explanation:

4 0
2 years ago
Precalc - trig identities. Decent explanation, please. THX!!!
LuckyWell [14K]

12. Recall the half-angle identity,

\sin^2\dfrac y2=\dfrac{1-\cos y}2\implies\sin\dfrac y2=\sqrt{\dfrac{1-\cos y}2}

where we take the positive square root because y is an angle in a right triangle, which means 0^\circ, so 0^\circ. For such an angle, it's always the case that \sin\dfrac y2>0.

Use the Pythagorean theorem to find the length of the hypotenuse:

\sqrt{5^2+12^2}=\sqrt{169}=13

Then

\sin\dfrac y2=\sqrt{\dfrac{1-\frac5{13}}2}=\sqrt{\dfrac4{13}}=\dfrac2{\sqrt{13}}

###

13. x is in quadrant II, which means 90^\circ, so 45^\circ. In other words, \dfrac x2 is in quadrant I, so \sin\dfrac x2>0. From the half-angle identity we get

\sin\dfrac x2=\sqrt{\dfrac{1-\cos x}2}=\sqrt{\dfrac23}

###

14. Simplification follows from the definitions of each function:

\sec x=\dfrac1{\cos x}

\tan x=\dfrac{\sin x}{\cos x}

\csc x=\dfrac1{\sin x}

So we have

\sec x\tan x\cos x\csc x=\dfrac{\sin x\cos x}{\cos^2x\sin x}=\dfrac1{\cos x}

###

15. Use the Pythagorean identity:

\cos^2x+\sin^2x=1\implies\dfrac{\cos^2x}{\cos^2x}+\dfrac{\sin^2x}{\cos^2x}=\dfrac1{\cos^2x}\implies1+\tan^2x=\sec^2x

Then

\tan^3x+\tan x=\tan x(\tan^2x+1)=\tan x\sec^2x

6 0
3 years ago
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