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2 years ago

Please can someone help thank you

Mathematics

1 answer:

Nitella [24]2 years ago

5 0

Answer:

it would be 36-45 and 56-64

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Which point is a solution to the inequality

julsineya [31]

HELLO! HOPE YOURE HAVING A GREAT DAY

your answer is C!

8 0

3 years ago

Is it an identity? and how?

frutty [35]

Yes, it's an identity because remember sin 6x would be (3x + 3x)= which would make it 2 sin 3x....concluding the equation should be an identity. :)

Hoped this helped :)

6 0

3 years ago

Write in simplest form:<br> <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24a%5E%7B10%7Db%5E%7B6%7D%7D" id="TexFormula1" ti

Charra [1.4K]

Answer:

$2a^3b^2\sqrt[3]{3a}$

Step-by-step explanation:

Use the following rules for exponents:

$a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x$

Simplify 24. Find two factors of 24, one of which should be a perfect cube:

$8*3=24\\\\2^3=8$

Insert:

$\sqrt[3]{2^3*3a^{10}b^6}$

Now split the exponents. Split 10 into as many 3's as possible:

$10=3+3+3+1$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}$

Split 6 into as many 3's as possible:

$6=3+3$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}$

Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):

$2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}$

Simplify:

$2a^3b^2\sqrt[3]{3a}$

:Done

6 0

2 years ago

If 0° ≤ θ ≤ 90° and cosθ = 11/15 , what is the value of sin (90° - θ)?

igor_vitrenko [27]

$11/15=0.7333$
$cos^{-1}(0.7333) =42.83342$
$sin(90-47.166572) = 0.7333$

first, find the numeric value for 11/15
second to find theta, simply do the <em>inverse</em> cos (which is cos^-1) of the first answer.
now you know theta, just do the sin of 90 - theta and that's it!

since you know whatr cos(theta) is, you just take the inverse cos of that number to get theta and 'reverse' cos, essentially. you are just solving for theta, by reversing the cos function with cos^-1
please mark as brainliest!

7 0

3 years ago

Read 2 more answers

Solve the equation.

Nutka1998 [239]

Answer:

Step-by-step explanation:

In this technique, if we have to factorise an expression like ax2+bx+c, we need to think of 2 numbers such that:

N1⋅N2=a⋅c=1⋅−12=−12

AND

N1+N2=b=−1

After trying out a few numbers we get N1=3 and N2=−4

3⋅−4=−12, and 3+(−4)=−1

x2−x−12=x2−4x+3x−12

x(x−4)+3(x−4)=0

(x+3)(x−4)=0

Now we equate the factors to zero.

x+3=0,x=−3

x−4=0,x=4

3 0

2 years ago