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3 years ago

Show all your working.

Mathematics

1 answer:

Wewaii [24]3 years ago

8 0

Answer:

The gravitational force acting between the two bodies is given by:

F=G

The dimension of the force is [MLT

−2

]

[M][M]

[MLT

−2

][L

]

−1

−2

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4) Expand and simplify<br> 5(x-3) - 3 (x + 5)

Nezavi [6.7K]

Answer:

2x-30

Step-by-step explanation:

5(x-3) = 5x-15

-3(x+5) -3x-15

(5x-15) + (-3x-15) = 2x-30

7 0

1 year ago

Ethan invests £500 in the bank for 2 years he earned £40 simple interest

Natalka [10]

Answer:

Rate = 4%

Step-by-step explanation:

principal = £500

time = 2 years

simple interest = £40

rate = ?

simple interest = PTR / 100

£40 = £500*2*R /100

£40*100 = £1000*R

£4000/£1000 = R

4 % = R

7 0

3 years ago

Solve and graph the solutions of the equation 2|x-2|-12=0

defon

2|x-2|-12=0

Rearrange variables to the left side of the question

2|x-2|=12

Dividing both sides of the equation by the coefficent of variables

|x-2|=12/2

Reduce the fraction

|x-2|=6

Split into two variables

x-2=6 or x-2=-6

Rearrange variables to the left side of the equation

x=6+2

x=8 or x=-4

when x=8

LHS=2|8-2|-12=0

RHS=0

LHS=RHS

x=8 is valid

A variable in mathematics is an alphabet or phrase that stands in for an unknowable amount, unknowable value, or unknowable number. Particularly in the case of algebraic expressions or algebra, the variables are utilised. As an illustration, the linear equation x+9=4 has the variables x and 9 as constants.

A variable is a letter or symbol that stands in for any individual integer in a group of two or more. A constant is a letter or symbol that stands for a single particular number, whether it is known or unknown.

In mathematics, variables serve as stand-ins for unknowable numbers. Variables can be utilized in formulae, equations, and expressions in mathematics. When it's necessary to find the optimum solution, variables are frequently employed.

Learn more about Variables brainly.com/question/17344045

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6 0

1 year ago

The equation of a circle is x2 − 4x + y2 + 8y − 29 = 0. what are the center and radius of the circle? enter your answers in the

uranmaximum [27]

$\text{The standard form of the circle:}\\\\(x-h)^2+(y-k)^2=r^2\\\\\text{h and k are the x and y coordinates of the circle}$

$\text{We have}\\\\x^2-4x+y^2+8y-29=0\\\\\text{use:(*)}(a\pm b)^2=a^2\pm2ab+b^2$
$x^2-2x\cdot2+y^2+2y\cdot4=29\ \ \ |+2^2\ \ |+4^2\\\\\underbrace{x^2-2x\cdot2+2^2}_{(*)}+\underbrace{y^2+2y\cdot4+4^2}_{(*)}=29+2^2+4^2\\\\(x-2)^2+(y+4)^2=49$
$\text{Answer: the center (2;-4), the radius}\ r=\sqrt{49}=7$

3 0

3 years ago

Please answer this question correctly and i'll mark you as brainlilest! Thank you.

raketka [301]

Answer:

$\frac{ {7}^{4} ( - 2) {}^{4} {11}^{3} 6}{7 {}^{3} {2}^{2} {6}^{2} 11 {}^{2} } = \frac{7 \times {2}^{2} \times 11}{6} = \frac{308}{6} = \frac{154}{3} \\ \\ \frac{ {7}^{0} {12}^{0} {17}^{0} }{ {57}^{0} + ( {42}^{0} \times ( {13}^{0} ))} = \frac{1}{1 + 1} = \frac{ 1}{2}$

4 0

3 years ago