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3 years ago

Any one that knows this please help thanks!

Mathematics

1 answer:

Vesnalui [34]3 years ago

3 0

If x = 4, you would replace the “x”s with 4’s! So the problem would work out to be 4^2 - 4 over 4 - 8. 4^2 = 16. You would then subtract. 16 - 4 = 12. Now your problem is 12 over 4 - 8. 4 - 8 = -4. Now your problem is 12 over -4. So now you just divide. 12 / -4 = -3! -3 should be your answer! If you need me to write it out to show you I’d be happy to! Hope this helps! :)

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2x+8y=6, -5x-20y=-15<br><br>need work because I don't understand

Vlada [557]

Answer:

Infinitely many solutions.

Step-by-step explanation:

Step: Solve 2x + 8y = 6 for x:

2x + 8y = 6

2x + 8y + −8y =6 + −8y (Add -8y to both sides)

2x = −8y + 6

2x/2 = (-8y + 6)/2 (Divide both sides by 2)

x = −4y + 3

Step: Substitute −4y + 3 for x in −5x − 20y = −15:

−5x − 20y = −15

−5 (−4y + 3 ) −20y = −15

−15 = −15 (Simplify both sides of the equation)

−15 + 15 = −15 + 15 (Add 15 to both sides)

0 = 0

4 0

3 years ago

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

The current temperature is 20 degrees. This is 6 degrees less than twice the temperature that it was 6 hours ago. What was the t

mestny [16]

2x - 6 = 20
2x = 20 + 6
2x = 26
x = 26/2
x = 13 <==

5 0

3 years ago

Arrange 3/5,5/8,5/6 and 7/4 in ascending order

melomori [17]

It’s already in ascending order.

3/5= .6

5/8= .625

5/6= .833

7/4= 1.75

6 0

2 years ago

A certain number of sixes and nines is added to give a sum of 126; if the number of sixes and nines is interchanged, the new sum

galina1969 [7]

Answer:

Original number of sixes = 6

Original number of nines = 10

Step-by-step explanation:

We are told in the question that:

A certain number of sixes and nines is added to give a sum of 126

Let us represent originally

the number of sixes = a

the number of nines = b

Hence:

6 × a + 9 × b = 126

6a + 9b = 126.....Equation 1

If the number of sixes and nines is interchanged, the new sum is 114.

For this second part, because it is interchanged,

Let us represent

the number of sixes = b

the number of nines = a

6 × b + 9 × a = 114

6b + 9a = 114.......Equation 2

9a + 6b = 114 .......Equation 2

6a + 9b = 126.....Equation 1

9a + 6b = 114 .......Equation 2

We solve using Elimination method

Multiply Equation 1 by the coefficient of a in Equation 2

Multiply Equation 2 by the coefficient of a in Equation 1

6a + 9b = 126.....Equation 1 × 9

9a + 6b = 114 .......Equation 2 × 6

54a + 81b = 1134 ........ Equation 3

54a + 36b = 684.........Equation 4

Subtract Equation 4 from Equation 3

= 45b = 450

divide both sides by b

45b/45 = 450/45

b = 10

Therefore, since the original the number of nines = b,

Original number of nines = 10

Also, to find the original number of sixes = a

We substitute 10 for b in Equation 1

6a + 9b = 126.....Equation 1

6a + 9 × 10 = 126

6a + 90 = 126

6a = 126 - 90

6a = 36

a = 36/6

a = 6

Therefore, the original number of sixes is 6

7 0

3 years ago