7 - 3x = 2x + 7 – 5x
Step by step

Answer:

2

Step-by-step explanation:

1+1×5−6+8×(4+4)−62 =

1+1x5-6+8x(8)-62 =

1+5-6+64-62 =

2

Hope this helps! :)

Answer:

Washington had 115 hits and Sanchez had 111 hits.

Step-by-step explanation:

2s + 4 = 226

2s = 222

s = 111

s + 4 = 115

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

{6, 8, 10, 12, 14, 16, 18, 20, ...}

Step-by-step explanation:

even natural numbers >/= 6.

Thus this has no upper limit and the set is thus infinite.

The set of even natural numbers greater than or equal to 6 is

" The natural numbers are those numbers used for counting and ordering. Numbers used for counting are called cardinal numbers."

We know the natural numbers are:

{1, 2, 3, 4, 5, 6, 7,..........}

Now, the set of even natural numbers are:

{2, 4, 6, 8, 10...........}

Therefore, the set of even natural numbers greater than or equal to 6 is

{6, 8, 10, 12,..............}

Learn more about natural numbers here: https://brainly.com/question/8102087

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Answer:

1833 miles is the total distance if the model airplane.

Answer:

D

Step-by-step explanation:

I hope thats correct

A glucose molecule is the smallest type of protein monomer of starch and cellulose largest of the carbohydrate polymers fundamental unit of genetic material.

A glucose molecule is the monomer of starch and cellulose.

Step-by-step explanation:

So from this, we can conclude that A glucose molecule is the monomer of starch and cellulose.

Learn more about glucose here:

brainly.com/question/13593795?referrer=searchResults

brainly.com/question/13159416?referrer=searchResults

Step-by-step explanation:

Please mark me as brainlyest

Answer:

-5-4i is your answer.

Answer:

3.42

Step-by-step explanation:

24/7=3.42

Answer:

3 eighths are needes to make up the rest of the measure.

Step-by-step explanation:

1/4 + 1/4 + 1/ 8 + x = 1

2/4 + 1/8+x=1

4/8 + 1/8+x=1

5/8 + x=1

x=1 - 5/8

x=3/8

We need three eights.(Because 3/8=1/8+1/8+1/8 and 1/8 represents an eights)

Answer:

C is the answer to your question

Answer:

g(f(x))=x^2 + 14x + 48

Step-by-step explanation:

you insert (x+7) into every x in the g(x) equation, so g(x)=(x+7)^2 - 1. then you simplify it by doing (x+7)(x+7) which gets you x^2 + 14x + 49. then you subtract the 1 from the equation to get g(f(x))=x^2 + 14x + 48.

(g o f) (x) = g ( f(x) )

= (x+7)² – 1

= x² + 14x+49 – 1

= x² + 14x+48

If you want, (f o g ) (x) below is the solution ^_^

(f o g ) (x) = f ( g(x) )

= g(x) + 7

= x²-1 +7

= x² +6

I hope I helped you^_^

If these vectors are to be linearly independent, then there exists some not-all-zero choice for scalars [tex]c_1,c_2,c_3,c_4[/tex] such that

[tex]c_1(1,1,0,-1) + c_2(1,k,1,1) + c_3(5,1,k,1) + c_4(-1,1,1,k) = (0,0,0,0)[/tex]

We can recast this as a system of linear equations,

[tex]\begin{bmatrix}1&1&5&-1\\1&k&1&1\\0&1&k&1\\-1&1&1&k\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\\c_4\end{bmatrix} = \begin{bmatrix}0\\0\\0\\0\end{bmatrix}[/tex]

This system has a solution if the coefficient matrix on the left side is not singular. So we attempt to find k such that it *is* singular, i.e. the determinant is zero:

• A cofactor expansion along the first column gives

[tex]\begin{vmatrix}1&1&5&-1\\1&k&1&1\\0&1&k&1\\-1&1&1&k\end{vmatrix} = \begin{vmatrix}k&1&1\\1&k&1\\1&1&k\end{vmatrix} - \begin{vmatrix}1&5&-1\\1&k&1\\1&1&k\end{vmatrix} + \begin{vmatrix}1&5&-1\\k&1&1\\1&k&1\end{vmatrix}[/tex]

• Cofactor expansions along the first columns of the remaining 3x3 matrices give

[tex]\begin{vmatrix}k&1&1\\1&k&1\\1&1&k\end{vmatrix} = k \begin{vmatrix}k&1\\1&k\end{vmatrix} - \begin{vmatrix}1&1\\1&k\end{vmatrix} + \begin{vmatrix}1&1\\k&1\end{vmatrix} = k^3-3k+2[/tex]

[tex]\begin{vmatrix}1&5&-1\\1&k&1\\1&1&k\end{vmatrix} = \begin{vmatrix}k&1\\1&k\end{vmatrix} - \begin{vmatrix}5&-1\\1&k\end{vmatrix} + \begin{vmatrix}5&-1\\k&1\end{vmatrix} = k^2-4k+3[/tex]

[tex]\begin{vmatrix}1&5&-1\\k&1&1\\1&k&1\end{vmatrix} = \begin{vmatrix}1&1\\k&1\end{vmatrix} - \begin{vmatrix}5&-1\\k&1\end{vmatrix} + \begin{vmatrix}5&-1\\1&1\end{vmatrix} = -k^2 - 6k + 7[/tex]

It follows that

[tex]\begin{vmatrix}1&1&5&-1\\1&k&1&1\\0&1&k&1\\-1&1&1&k\end{vmatrix} = (k^3-3k+2) - (k^2-4k+3) + (-k^2 - 6k + 7) \\\\ \begin{vmatrix}\cdots\end{vmatrix} = k^3 - 2k^2 - 5k + 6 = (k-3)(k-1)(k+2)[/tex]

which makes the coefficient matrix singular if k = 3, k = 1, or k = -2.

Then the four vectors are linearly independent for

[tex]\left\{ k\in\mathbb R \mid k\not\in\{-2,1,3\}\right\}[/tex]

Answer:

2x+1=0.

2x=0-1.

2x=-1.

2x/2=-1/2.

x=-2.

3x+4=0

3x=0-4

3x=-4

3x/3=-4/3

[tex]x = 1 \frac{1 }{3} [/tex]

Answer:

p = k(d)/u^2

Step-by-step explanation:

BRAINLIEST, PLEASE!

Answer:

8.5 feet and 7.5 feet.

Step-by-step explanation:

Hope this helps :)

The guitar's volume was amplified, but it was hard to hear the notes since it was so loud.

What word or phrase is a signal word that a context clue might be nearby to help determine the meaning of "amplified"?

Step-by-step explanation:

9514 1404 393

Answer:

  p = -3/4b -3/4

Step-by-step explanation:

Vectors are perpendicular when their dot product is zero.

  A·B = p(4) +b(3) -3(-1) = 4p +3b +3

We want this to be zero, so ...

  4p +3b +3 = 0

  4p = -3b -3

  p = -3/4b -3/4 . . . . the value of p that makes A⊥B

Answer:

1/4

Step-by-step explanation:

Here we want to see if the statement:

"A proper use of units can help us to determine the meaning of the slope of a linear function" is true or false.

We will see that it is true.

First, a general linear equation can be written as:

y = a*x + b

Where a is the slope, x is the independent variable, y is the dependent variable, and b is the y-intercept.

The proper use of units is:

y and b must have the same units.

x can have any type of units

The slope must have the quotient between the units of y and the units of x.

Where the proper use of units means that we have the same units on both sides of the expression.

Now, assume that we know the units of y and x.

For example:

y is miles  (then b is also in miles)

x is in hours

Then, with the proper use of units, we can see that the units of the slope are:

miles/hour, or miles per hour.

With this (only knowing the units of x and y) we can see that the slope will be a speed.

So yes, the proper use of units is really important, as this has a lot of information (particularly, the meaning of the slope in a linear equation)

Thus, the statement is true.

If you want to learn more, you can read:

https://brainly.com/question/20286983

Answer:

4.  3,52  - Tenths                

   0.352 - Hundredths

5.  65.09 - Ones

   7.005 - Thousandths

6.  753.4 - Tens

    0.051 -  Hundredths

Step-by-step explanation:

all steps are in picture

Answer:

[tex]{ \underline{ \sf{x \: is \: \: \{ \frac{ - 5 + i \sqrt{35} }{10} \} } \: \: and \: \: \{ \frac{ - 5 - i \sqrt{35} }{10} \}} }[/tex]

Step-by-step explanation:

[tex]5 {x}^{2} + 4x + 3 = 0 [/tex]

from quadratic formular:

[tex]{ \sf{x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} }} [/tex]

general equation:

[tex] {ax}^{2} + bx + c = 0[/tex]

a is 5, b is 4 and c is 3:

substitute in formular:

[tex]{ \sf{x = \frac{ - 5± \sqrt{ {5}^{2} - (4 \times 5 \times 3) } }{(2 \times 5)} }} \\ \\ { \sf{x = \frac{ - 5± \sqrt{ - 35} }{10} }}[/tex]

but from complexes, i² = -1

[tex]{ \sf{x = \frac{ - 5± \sqrt{35 {i}^{2} } }{10} }} \\ \\ = { \sf{x = \frac{ - 5±i \sqrt{35} }{10} }}[/tex]

Step-by-step explanation:

11.

[tex] \frac{x - 4}{ \frac{x}{4} - \frac{4}{x} } [/tex]

[tex] \frac{x - 4}{ \frac{ {x}^{2} }{4x} - \frac{16}{4x} } [/tex]

[tex] \frac{x - 4}{ \frac{ {x}^{2} - 16 }{4x} } [/tex]

[tex] \frac{x - 4}{ \frac{(x + 4)(x - 4)}{4x} } [/tex]

[tex] \frac{4x}{x + 4} [/tex]

12.

[tex] \frac{ \frac{ {x}^{2} - 1 }{x} }{ \frac{(x - 1) {}^{2} }{x} } [/tex]

[tex] \frac{ {x}^{3} - x}{ {x}^{3} - 2 {x}^{2} + x } [/tex]

[tex] \frac{x( {x} - 1)(x + 1) }{x( {x} - 1) {}(x - 1) } [/tex]

[tex] \frac{x + 1}{x - 1} [/tex]

13.

[tex] \frac{ \frac{ {t}^{2} }{ \sqrt{ {t}^{2} + 1} } - \sqrt{ {t}^{2} + 1} }{ {t}^{2} } [/tex]

[tex] \frac{ \frac{ {t}^{2} }{ \sqrt{ {t}^{2} + 1 } } - \frac{ {t}^{2} + 1}{ \sqrt{ {t}^{2} + 1 } } }{ {t}^{2} } [/tex]

[tex] \frac{ \frac{1}{ \sqrt{ {t}^{2} + 1 } } }{ {t}^{2} } [/tex]

[tex] \frac{ - \frac{ \sqrt{ {t}^{2} + 1 } }{ {t}^{2} + 1 } }{ {t}^{2} } [/tex]

[tex] - \frac{ \sqrt{ {t}^{2} + 1} }{ {t}^{2}( {t}^{2} + 1) } [/tex]

14.

[tex] \frac{3x { }^{ \frac{1}{3} } - x {}^{ \frac{ - 2}{3} } }{3x {}^{ - \frac{2}{3} } } [/tex]

[tex] \frac{3x - 1}{3} [/tex]