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2 years ago

Which angles are neither obtuse angles nor acute angles?

Mathematics

2 answers:

sleet_krkn [62]2 years ago

5 0

Answer:

A right angle is neither Obtuse nor Acute

Step-by-step explanation:

Akimi4 [234]2 years ago

3 0

Answer: Which angles are neither obtuse angles nor acute angles?

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Help plz:)))I’ll mark u Brainliest

zheka24 [161]

Answer: The answer is 10

3 0

3 years ago

Jerry has a jar with 273 marbles in it. The jar can hold 1,000 marbles in all

Montano1993 [528]

Answer:

I would say that the answer is 1000=273+m

7 0

3 years ago

$7000 is compounded semiannually at a rate of 11% for 21years Find the total amount and round to the nearest cent

photoshop1234 [79]

$\text{Given that }\$7000 \text{ is compounded semiannualy at a rate of }11\% \text{ for 21 years}\\
\\
\text{we know that the amount after t year when compounded is given by}\\
\\
A=P\left ( 1+\frac{r}{n} \right )^{nt}\\
\\
\text{here P is the principal amoount, so }P=7000,\\
\\
\text{r is the interest rate, }r=11\%=0.11\\
\\
\text{n is the number of times in a year, here semiannulay, so }n=2,\\
\\
\text{and t is the time, so }t=21\\
\\
\text{so the amount after 21 years is}$

$A=7000\left ( 1+\frac{0.11}{2} \right )^{2(21)}\\
\\
\Rightarrow A=7000\left ( 1+0.055 \right )^{42}\\
\\
\Rightarrow A=7000\left ( 1.055 \right )^{42}\\
\\
\Rightarrow A\approx 66328.68$

So the amount after 21 years is: $66328.68

7 0

3 years ago

72 + 8x + 18x = 36x

ArbitrLikvidat [17]

$72 + 8x + 18x = 36x
$

$72+26x=36x
$

$72=36x-26x
$ $\text{Subtract 26x from both sides}$

$72=10x
$

$\text{Divide both sides by 10}$

$x=36/5$

4 0

3 years ago

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

8 0

3 years ago