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3 years ago

Please HELPP!!! Solve the system of equations. y= 2x y= x^2 - 15

1 answer:

7 0

$y=2x\\y=x^2-15\\\\x^2-15=2x\\x^2-2x-15=0\\x^2-5x+3x-15=0\\x(x-5)+3(x-5)=0\\(x+3)(x-5)=0\\x=-3 \vee x=5\\\\y=2\cdot(-3) \vee y=2\cdot5\\y=-6 \vee y=10\\\\(x,y)\in\{(-3,-6),(5,10)\}$

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Which expression is equivalent to 30x + 15y?

Softa [21]

C) because you can divide both by 15 and 30÷15 = 2

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3 years ago

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The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (

kow [346]

Answer:

$\|a\| = 5\sqrt{13}$ .

$\|b\| = 3\sqrt{29}$ .

Step-by-step explanation:

Let $m$ , $n$ , and $k$ be scalars such that:

$\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}$ .

$\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}$ .

$\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}$ .

The question states that $\| a + b \| = 34$ . In other words:

$k\, \sqrt{8^{2} + 15^{2}} = 34$ .

$k^{2} \, (8^{2} + 15^{2}) = 34^{2}$ .

$289\, k^{2} = 34^{2}$ .

Make use of the fact that $289 = 17^{2}$ whereas $34 = 2 \times 17$ .

$\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}$ .

$k^{2} = 2^{2}$ .

The question also states that the scalar multiple here is positive. Hence, $k = 2$ .

Therefore:

$\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}$ .

$(a + b)$ could also be expressed in terms of $m$ and $n$ :

$\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}$ .

$\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}$ .

Equate the two expressions and solve for $m$ and $n$ :

$\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}$ .

$\begin{cases}m = 5 \\ n = 3\end{cases}$ .

Hence:

$\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}$ .

$\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}$ .

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3 years ago

this isn’t a question but i just wanted to tell everyone you all are doing great. this year has been hard and everyone is workin

BlackZzzverrR [31]

Thank you so much friend.

I think a advantage of this will be that we will be able to tell our grandchildren about how we survived the pandemic, just like our grandparents talk about World War. Lol.

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2 years ago

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Which would be the next step in solving for matrix A?

Lady_Fox [76]

Answer:

A=(16 -5 0 -9)-(3 -4 -11 24)

Step-by-step explanation:

The solution process for any algebraic expression is to "undo" what is done to the variable. Here, matrix A has (3 -4 -11 24) added to it. The next step is to undo that addition, by subtracting that amount from both sides of the equation. The result is ...

A = (16 -5 0 -9)-(3 -4 -11 24)

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We assume there are typos in the answer choices listed here.

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madreJ [45]

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2 years ago

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