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2 years ago

-2(5y - 8) help me now

Mathematics

2 answers:

hammer [34]2 years ago

7 0

Answer:

-10y +16 (if this isn't right tell me exactly what you have to do)

sweet-ann [11.9K]2 years ago

3 0

Answer:

-2×5y-2×(-8)=

= -10y+16

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Expanded form for 1.7

Verdich [7]

One and seven tenths

7 0

2 years ago

Read 2 more answers

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:

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

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

6 0

3 years ago

One zero of x3 – 3x2 – 6x + 8 = 0 is -2. What are the other zeros of the function?

hram777 [196]

Consider this solution, if it is possible, check the result:
1) if to divide polynomial 'x³-3x²-6x+8' on 'x+2' result is: x²-5x+4, then x³-3x²-6x+8=(x+2)(x²-5x+4);
2) x²-5x+4=(x-1)(x-4), then
3) x³-3x²-6x+8=(x+2)(x-1)(x-4).
4) zeros are: -2;1;4. Other zeros are: 1 and 4.
Answer: 1;4.

7 0

3 years ago

Write the expression 12 x 4^2 in words

lawyer [7]

Answer and Step-by-step explanation:

Twelve times four squared.

This is how you would write the expression in word format.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

7 0

2 years ago

I need help with 2 and 5 and I need you to right down how you did it plz and thank you

anastassius [24]

Number 5 is 22 and 35
For example: 7+1=8, 8+2 =10, 10+3=13, 13+4=17, 17+5=22, 22+6=28, 28+7=35

5 0

2 years ago