<img src="https://tex.z-dn.net/?f=x%5E2-11x%2B24%3D0" id="TexFormula1" title="x^2-11x+24=0" alt="x^2-11x+24=0" align="absmiddle"

because there are 4 people in total including Mrs Cats you would add up 33 + 17 + 3 + 6 to get 59 then you would divide that by the amount of people there are in this case 4 to get 14.75.

5 0

2 years ago

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(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

8 0

2 years ago

Verifying Properties of logarithms In Exercise,use a graphing utility to verify that the functions are equivalent for x > 0.

Ede4ka [16]

Answer:

Applying Quotient and power rule. Graphically, for x>0 both curves coincide.

Step-by-step explanation:

1) Firstly, by applying the Quotient and then the Power Rule, we have a subtraction of the argument and, finally the exponent turns to be the coefficient. As it follows:

$lnx=log_{e}x\\\\f(x) = ln({\frac{x^{2}}{4}})\Rightarrow f(x)=ln{x^2}-ln4\Rightarrow f(x)=2lnx-ln4\\g(x) = 2 ln x - ln 4$

2) Check the graph below to see this equivalence. to see x>0 they both coincide.

7 0

2 years ago