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3 years ago

The value of the 7 in 27,459 is how many times the value of the 7 in 40,735?

Mathematics

2 answers:

Digiron [165]3 years ago

7 0

Answer:

The answer: 10

alukav5142 [94]3 years ago

4 0

Answer:

Step-by-step explanation:

The 7 in 27,459 is in the thousands place.

The 7 in 40,735 is in the hundreds place.

Any digit in the thousands place has 10 times the value of the same digit in the hundreds place. (The multiplier is 1000, instead of 100.)

_____

In the base-10 number system, each digit location has 10 times the value of the location to its right. The value is determined by the place the digit occupies in the number. That is why it is called a place-value number system.

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A triangle has base 2x+1 and height 6x-3. What value of x would give an area of 240m2?

GREYUIT [131]

Answer:

The values of x which would give an area of 240m² would be:

$x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}$

Step-by-step explanation:

Given

The base of triangle b = 2x+1

The height of triangle h = 6x-3

The Area of the triangle A = 240 m²

The Area of the triangle has the formula

A = 1/2 × b × h

substituting b = 2x+1, h = 6x-3 and A = 240

$240\:=\:\frac{1}{2}\:\left(2x+1\right)\:\times \left(6x-3\right)$

$480=\left(2x+1\right)\left(6x-3\right)$

$480=12x^2-3$

Subtract 480 from both sides

$12x^2-3-480=480-480$

$12x^2-483=0$

$3\left(4x^2-161\right)=0$

$3\left(2x+\sqrt{161}\right)\left(2x-\sqrt{161}\right)=0$

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

$2x+\sqrt{161}=0\quad \mathrm{or}\quad \:2x-\sqrt{161}=0$

solving

$2x+\sqrt{161}=0$

$2x=-\sqrt{161}$

Divide both sides by 2

$\frac{2x}{2}=\frac{-\sqrt{161}}{2}$

$x=-\frac{\sqrt{161}}{2}$

also solving

$2x-\sqrt{161}=0$

$2x=\sqrt{161}$

Divide both sides by 2

$\frac{2x}{2}=\frac{\sqrt{161}}{2}$

$x=\frac{\sqrt{161}}{2}$

Therefore, the values of x which would give an area of 240m² would be:

$x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}$

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2 years ago

What is the 2nd equivalent fraction to 1/11

alisha [4.7K]

2/22.
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2 years ago

Simplify √49 + [√81 - x(9x = 14)]

eimsori [14]

$\longrightarrow{\green{- 9 {x}^{2} + 14x + 16}}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}$

$\sqrt{49} + [ \sqrt{81} - x \: (9x - 14) ] \\ \\ = \sqrt{7 \times 7} + [ \sqrt{9 \times 9} - 9 {x}^{2} + 14x] \\ \\ = \sqrt{( {7})^{2} } + [ \sqrt{ ({9})^{2} } - 9 {x}^{2} + 14x ] \\ \\ (∵ \sqrt{ ({x})^{2} } = x ) \\ \\ = 7 + (9 - 9 {x}^{2} + 14x) \\ \\ = 7 + 9 - 9 {x}^{2} + 14x \\ \\ = - 9 {x}^{2} + 14x + 16$

$\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{♡}}}}}$

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3 years ago