A basketball player made 80 out of 100 attempted free throws what percent of free throws was made ?

Mathematics

2 answers:

AveGali [126]2 years ago

8 0

Answer:

80%

Step-by-step explanation:

80÷100=.8 which is 80%

Tom [10]2 years ago

6 0

Answer:

8/10

Step-by-step explanation:

id hope your have a great day

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Plz find the iii (3).plz plz plz

kirza4 [7]

Answer:

Rs 937.50

Step-by-step explanation:

Rs 750 for 4 years at 6 1/4% interest:

Simple interest is found using i = p*r*t, where p is the principal, r is the interest rate as a decimal fraction, and t is the number of years.

Here, i = (Rs 750)*0.0625*4 = Rs 187.50

The amount (that is, principal plus simple interest) will be:

Rs 750 + Rs 187.50 = Rs 937.50

8 0

2 years ago

Please please help!!

shepuryov [24]

Answer:

$\large\boxed{x=0\ and\ x=\pi}$

Step-by-step explanation:

$\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2$

$\dfrac{\sin^2x}{(\cos^2x)^2}+\dfrac{2-\sin^2x}{\cos^2x}=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-(1-\cos^2x)}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-1+\cos^2x}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{1+\cos^2x}{\cos^2x}=2$

$\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x$