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3 years ago

Write the following as a complex number in the form a + bi square root -245

Mathematics

2 answers:

Pepsi [2]3 years ago

5 0

Answer:

Step-by-step explanation:

√(-245)=√(49×5×i^2)=7√5 i=0+7√5 i

svetoff [14.1K]3 years ago

3 0

Answer:

$\sqrt{-245}$ in the form a + bi is $0+7\sqrt{5}i$

Step-by-step explanation:

We know that

$i=\sqrt{-1}$

We need to find $\sqrt{-245}$

$\sqrt{-245}=\sqrt{-1}\times \sqrt{245}=i\sqrt{245}=i\sqrt{49\times 5}=i\times 7\sqrt{5}$

$\sqrt{-245}$ in the form a + bi is $0+7\sqrt{5}i$

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what is the scale factor of triangle UVW to triangle XYZ where uvw has side Lengths 3,4,2, and XYZ has lengths 14,28,21

alekssr [168]

Answer: 7

Step-by-step explanation:

Given: Triangle UVW map to triangle XYZ where Triangle UVWhas side Lengths 3,4,2, and triangle XYZ has lengths 14,28,21.

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3 years ago

-5x - 8y = 17<br> 2x - 7y = -17

Mrac [35]

Answer:

x = -5

y = 1

Step-by-step explanation:

-5x - 8y = 17

x = -5

y = 1

2x - 7y = -17

x = -5

y = 1

4 0

3 years ago

Please help

Alekssandra [29.7K]

Answer:

25 in x 15 in

Step-by-step explanation:

Given:

Length = 3/5 the width
Area = 375 in²

Let width = $x$

Therefore, length = 3/5 $x$

First create an equation for the area of the picture based on the given information for its width and length:

$\begin{aligned} \implies \textsf{Area of original picture} & = \sf width \times length\\& = x\left(\dfrac{3}{5}x\right)\\& = \dfrac{3}{5}x^2\end{aligned}$

We are told the area of the enlarged picture is 375 in². Therefore, substitute this into the equation and solve for $x$ to find the width of the enlarged picture:

$\begin{aligned}\textsf{Area} & = 375\\ \implies \dfrac{3}{5}x^2 & = 375\\ x^2 & =375 \cdot \dfrac{5}{3}\\ x^2 & =625\\ x & = \sqrt{625}\\ x& = 25\end{aligned}$

Therefore, the width of the enlarged picture is 25 in.

Substitute the found value of $x$ into the expression for length to find the length of the enlarged picture:

$\begin{aligned}\sf Length & = \dfrac{3}{5}x\\\\\implies \sf Length & = \dfrac{3}{5}(25)\\\\& = 15\: \sf in\end{aligned}$

Therefore, the dimensions of the enlarged picture are <u>25 in x 15 in</u>. The width is 25 in and the length is 15 in, as the length is 3/5 of the width.

5 0

2 years ago