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3 years ago

Write the relationship between the values of the given digits like 5s in 2,755

Mathematics

2 answers:

ladessa [460]3 years ago

7 0

Answer:

Value of second 5 is 10 times the value of first 5.

Step-by-step explanation:

Given : Number 2,755.

To find : Write the relationship between the values of the given digits like 5s .

Solution : We have given 2,755.

Here we can there are two 5's.

First 5 from the right us at ones place = 5

Second 5 from the right is at tens place = 50.

So, 50 = 10 * 5

Value of second 5 is 10 times the value of first 5.

Therefore, Value of second 5 is 10 times the value of first 5.

stealth61 [152]3 years ago

4 0

The first 5 is in the 10s place. The second is in the ones. To write this out, it would be 5 tens (50), plus 5 ones (5); 55. I don't know if this is what you were looking for, but I hope I could help in some way!

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Prove that tan 10+tan 70'+taniou = tanio tanfo'tam 1oo.

Galina-37 [17]

Question:

Prove that:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Answer:

Proved

Step-by-step explanation:

Given

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Required

Prove

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Subtract tan(10) from both sides

$- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

$tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

Factorize the right hand size

$tan(70) + tan(100) = -tan(10)(-tan(70). tan(100) + 1)$

Rewrite as:

$tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))$

Divide both sides by $1-tan(70). tan(100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$

In trigonometry:

$tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}$

So:

$\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)}$ can be expressed as: $tan(70 + 100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$ gives

$tan(70 + 100) = -tan(10)$

$tan(170) = -tan(10)$

In trigonometry:

$tan(180 - \theta) = -tan(\theta)$

So:

$tan(180 - 10) = -tan(10)$

Because RHS = LHS

Then:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$ has been proven

6 0

3 years ago

If the diagonals of a parallelogram from right angles,then the parallelogram is the rectangle

geniusboy [140]

The statement above is false.

If the diagonals of a parallelogram form right angles, then the parallelogram is a rhombus (a rhombus is a quadrilateral with four equal side lengths).

Note* = by saying the statement is false is not saying that the scenario presented in the statement cannot occur. If the rectangle was a square, then its diagonals can form right angles since a square is also a rhombus. However, if a rectangle was NOT a square, its diagonals would not form right angles. A true statement is a statement where ALL cases fit the said requirement(s).

The statement can also be corrected by saying:

If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.

All rectangles (even a square) have congruent diagonals, so this statement would be true.

Hope this helps!

8 0

3 years ago

Calculate y as a function of x when dy/dx = 4x3 + 3x2 - 6x + 5

PtichkaEL [24]

Answer:
y = x⁴ + x³ - 3x² + 5x + C

======

Separable differential equations such as these ones can be solved by treating dy/dx as a ratio of differentials. Then move the dx with all the x terms and move the dy with all the y terms. After that, integrate both sides of the equation.

$\begin{aligned}
\dfrac{dy}{dx} &= 4x^3 + 3x^2 - 6x + 5 \\
dy &= (4x^3 + 3x^2 - 6x + 5) dx \\
\int dy &= \int (4x^3 + 3x^2 - 6x + 5) dx 
\end{aligned}$

In general (understood that +C portions are still there),

$\int x^{m} = \dfrac{x^{m+1}}{m+1}$

Note that ∫dy = y since it is ∫1·dy = ∫y⁰ dy = y¹/(0+1) = y
For the right-hand side, we use the sum/difference rule for integrals, which says that

$\int \big[f(x) \pm g(x)\big]\, dx = \int f(x)\,dx \pm \int g(x) \, dx$

Applying these concepts:

$\begin{aligned} 
 \int dy &= \int (4x^3 + 3x^2 - 6x + 5) \, dx \\
y &= \int 4x^3\,dx + \int 3x^2 \, dx - \int 6x\, dx + \int 5\, dx \\
&= \frac{4x^4}{4} + \frac{3x^3}{3} - \frac{6x^2}{2} + 5x + C \qquad \text{(only one $C$ is needed)} \\
&= x^4 + x^3 - 3x^2 + 5x + C
\end{aligned}$

The answer is y = x⁴ + x³ - 3x² + 5x + C

6 0

3 years ago

A concrete patio is

astra-53 [7]

The length of concrete patio is 6.5 feet which is less than given 7 foot picnic table. So the concrete slab is not enough to fit a 7 foot picnic table

<em><u>Solution:</u></em>

Given that concrete patio is $5\frac{2}{3}$ feet wide

Area of concrete patio = $36\frac{2}{3}$ square feet

Therefore,

$\texttt{Width of concrete patio = }5\frac{2}{3}=\frac{5\times 3+2}{3}=\frac{17}{3}feet\\\\\text{Area of concrete patio = }36\frac{2}{3}=\frac{36\times 3+2}{3}=\frac{110}{3}feet^2$