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3 years ago

What is 90% in the simplest form?

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

7 0

90%=.90 .... 90/100 === 9/10 in simplest form

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One positive number is 9 more than twice another. If their product is 95, find the numbers.

mafiozo [28]

Answer:

The numbers are 19 and 5

Step-by-step explanation:

Given

Let the numbers be $x\ and\ y.$

So:

$x = 9 + 2y$ --- First statement

$x*y = 95$ --- second statement

Required

Find x and y

Substitute $x = 9 + 2y$ in $x*y = 95$

$(9 + 2y) * y = 95$

$9y + 2y^2 = 95$

Rewrite as:

$2y^2 + 9y - 95 = 0$

Expand

$2y^2 -10y +19y- 95 = 0$

Factorize

$2y(y -5) +19(y- 5) = 0$

$(2y +19)(y- 5) = 0$

Solve for y

$2y + 19 =0$ or $y - 5 = 0$

$2y = -19$ or $y = 5$

$y = -\frac{19}{2}$ or $y=5$

Since the numbers are positive, we take only:

$y=5$

Substitute $y=5$ in $x = 9 + 2y$

$x = 9 + 2 * 5$

$x = 9 + 10$

$x = 19$

The numbers are 19 and 5

5 0

3 years ago

PLEASE HELP 100 points to who ever gives me the answer Compare the numbers using >, <, or =.

never [62]

Answer:

a) -2 < 3

b) |-12| < |15| (the absolute value of -12 is 12 so that's comparing 12 to 15)

c) 3 > -4

d) |15| > |-12|

e) 7 > -11

7 0

3 years ago

This diagram is a straightedge and compass construction. A is the center of one circle, and B

Serhud [2]

Answer:

Correct choices are a, b, e

Step-by-step explanation:

As per diagram, the circles are equal.

Since the distance AB is the radius of both circles:

AB = AC = BC = BD = AD = r

So the answer choices:

a. AC = BC

Correct, both equal to r

b. AC = BD

Correct, both equal to r

C. CD = AB

Incorrect. Half of CD is the leg of 30-60-90 triangle and is equal to r√3/2, so CD = AB√3

d. ABCD is a square

Incorrect. ACBD is rhombus

e. ABD is an equilateral triangle

Correct, all three sides are equal to r

f. CD = AB + AB

Incorrect as CD = AB√3

3 0

3 years ago

Find the exact value of the trigonometric expression.

Dmitrij [34]

$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}$

$\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

7 0

4 years ago

An account earns simple interest. Find the annual interest rate.

vlada-n [284]

Annual interest rate is 225

4 0

3 years ago