All categories

2 years ago

Question 1/8

Mathematics

1 answer:

just olya [345]2 years ago

5 0

Answer:

discount = 290 × 1

= 72.50

new price is

290.00

72.50 -

217.50

You might be interested in

20 teachers selected sun rise beach. 12 teachers selected Mkapa bridge and St. Anthony's area. 16 teachers selected St.Anthony's

mart [117]

Answer:

vaq-kdae-dfm come all bad gi rls

8 0

2 years ago

WHAT IS THE VALUE OF x IN CENTIMETERS A. 30cm B. 8cm C. 10.8cm D.22.5cm

Katarina [22]

Answer:

D)22.5cm

CB/BA=HG/13.5

CB=15cm,BA=9cm,HG=x and GF=13.5

15/9=x/13.5(divided by 13.5both sides)

x=22.5cm

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cfrak%7B%20%5Corange%7B%20Question%7D%7D" id="TexFormula1" title=" \large\frak{ \

SVEN [57.7K]

Answer:

$\boxed{\sf distance \: between \: the \: objects \: \: is \: 2.5 \: cm}$

Step-by-step explanation:

According to universal law of gravitation every object in the universe attracts every other particles that surrounds it with a force which is inversely proportional to the square of their distance of separation & directly proportional to

the product of their masses, given by standard formula.

$A = G \cdot \frac{m_1.m_2}{ {d}^{2} }$

where A,d & G are force of attraction, distance of separation & proportionality constant respectively.

Given:

A1 = 2 units

d1= 5 cm

A2 = 8 units

To find:

Distance of separation when the force of attraction is 8 units d2 = ?

Solution:

Substituting the given values in above at each point,

$A_1 = G \cdot \frac{m_1.m_2}{ {d_1}^{2} } \\ 2 = G \cdot \frac{m_1.m_2}{ {5}^{2} } \\ \sf similarly \: at \: second \: point \: of \: attraction \\ A_2 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ 8 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ \sf \: \: dividing \: both \: of \: the \: equation \\ \frac{2}{8} = \frac{G \cdot \frac{m_1.m_2}{ {5}^{2} }}{G \cdot \frac{m_1.m_2}{ {d_2}^{2} } } \\ \sf \: G \: m_1 and \: m_2 \: are \: \: constant \: hence \\ \sf they \: can \: be \: cancelled \: out \\ \frac{2}{8} = \frac{ \frac{1}{ {5}^{2} } }{ \frac{1}{ {d_2}^{2} } } \\ \sf rearranging \: above \: equation \\ \frac{1}{4} = \frac{{d_2}^{2} }{ {5}^{2} } \\ {d_2}^{2} = \frac{ {5}^{2} }{4 } \\ {d_2}^{2} = \frac{25}{4} \\ {d_2}^{2} = 6.25 \\ \sqrt{{d_2}^{2} } = \sqrt{6.25} \\ \boxed{ \sf{d_2} = 2.5 \: cm}$

Answer: the distance between the two objects is 2.5 cm, if the attraction between them is 8 units.

Learn more about universal law of gravitation here brainly.com/question/27244479

Thanks for joining brainly community!

7 0

1 year ago

4/x-1+6/x+1=-12/x^2-1

Mademuasel [1]

$\frac{4}{x - 1} + \frac{6}{x + 1} = \frac{-12}{x^{2} - 1}$

$\frac{4(x + 1)}{(x - 1)(x + 1)} + \frac{6(x - 1)}{(x - 1)(x + 1)} = \frac{-12}{x^{2} - 1}$

$\frac{4x + 4}{x^{2} - 1} + \frac{6x - 6}{x^{2} - 1} = \frac{-12}{x^{2} - 1}$

$\frac{10x - 2}{x^{2} - 1} = \frac{-12}{x^{2} - 1}$

$(10x - 2)(x^{2} - 1) = -12(x^{2} - 1)$
$10x^{3} - 10x^{2} - 2x^{2} + 2 = -12(x^{2}) + 12(1)$
$10x^{3} - 12x^{2} + 2 = -12x^{2} + 12$
$10x^{3} - 12x^{2} + 12x^{2} + 2 = -12x^{2} + 12x^{2} + 12$
$10x^{3} + 2 - 2 = 12 - 2$
$\frac{10x^{3}}{10} = \frac{10}{10}$

$x^{3} = 1$
$x = 1$

4 0

2 years ago

Simplify the rational expressions. state any excluded values

dlinn [17]

Answer:

8.) 3

9.) x+5

10.) 50x/2x-5

11. 3n4-n3(cubed)/n3+n2-n-1

Step-by-step explanation:

4 0

2 years ago