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4 years ago

What is 3.47 rounded to the nearest tenth

Mathematics

2 answers:

ch4aika [34]4 years ago

4 0

3.5 is rounded to the nearest tenth.

Alex777 [14]4 years ago

3 0

The answer is 3.5
becuase u rounded up

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Michael's dad is 30 years of age. He is 2 years more than four times Michaels age m. Write and solve a two-step equation to dete

Ksenya-84 [330]

Answer:

The two step equation that we can use to find michael's age is x = (f-2)/4 where f = 30. So Michael is 7 years old.

Step-by-step explanation:

In order to solve this problem we will attribute variables to the ages of Michael and his father. For his father age we will attribute a variable called "f" and for Michael's age we will attribute a variable called "x". The first information that the problem gives us is that Michael's dad is 30 years of age, so we have:

f = 30

Then the problem states that the age of the father is 2 years "more" than four "times" Michaels age. The "more" implies a sum and the "times" implies a product, so we have:

f = 2 + 4*x

We can now find Michael's age, for that we need to isolate the "x" variable. We have:

f - 2 = 4*x

4*x = f - 2

x = (f-2)/4

x = (30 - 2)/4 = 7 years

The two step equation that we can use to find michael's age is x = (f-2)/4 where f = 30. So Michael is 7 years old.

7 0

4 years ago

Read 2 more answers

A rectangle has a side length of 2x and a diagonal of 8x. What is the missing lenght?

Leona [35]

Answer:

2x $\sqrt{15}$

Step-by-step explanation:

we can use the Pythagorean Theorem to find the missing side (a):

a² + (2x)² = (8x)²

a² + 4x² = 64x²

a² = 60x²

a = √60x²

a = $\sqrt{4}$ · $\sqrt{15}$ · $\sqrt{x^2}$

This can be simplified to be 2x $\sqrt{15}$

6 0

3 years ago

What does it mean when a number is out of the parenthesis? Example 3($28+$32)-$10=

777dan777 [17]

When the number is out of parenthesis it means it is multiplying the number between the parenthesis
example - 3 ( 28 + 32 ) => 3 ( 60)
which means 3 * 60 = 180

Note:- you cannot multiply the numbers when they are 3 ( 28 + 32 ) because you have to use PEMDAS ( order of operation ) to solve the problem where P = parenthesis , E = exponents, M = multiplication, D = divide , A = addition , S = subtract.

4 0

3 years ago

Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt

Paladinen [302]

Answer:

$y(x)=4a\sqrt{3}* sin(x)-3a$

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

$\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}$

But:

$\frac{1}{tan(x)} =cot(x)$

So:

$\frac{dy(x)}{dx} =cot(x)*(3a+y)$

Multiplying both sides by dx and dividing both sides by 3a+y:

$\frac{dy}{3a+y} =cot(x)dx$

Integrating both sides:

$\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx$

Evaluating the integrals:

$log(3a+y)=log(sin(x))+C_1$

Where C1 is an arbitrary constant.

Solving for y:

$y(x)=-3a+e^{C_1} sin(x)$

$e^{C_1} =constant$

So:

$y(x)=C_1*sin(x)-3a$

Finally, let's evaluate the initial condition in order to find C1:

$y(\frac{\pi}{3} )=3a=C_1*sin(\frac{\pi}{3})-3a\\ 3a=C_1*\frac{\sqrt{3} }{2} -3a$

Solving for C1:

$C_1=4a\sqrt{3}$

Therefore:

$y(x)=4a\sqrt{3}* sin(x)-3a$

3 0

4 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

3 years ago