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1 year ago

<>) Melinda

Mathematics

1 answer:

Vanyuwa [196]1 year ago

7 0

Answer:

The statement "500 divided by 50 is 10" is a true statement

Step-by-step explanation:

In mathematics, we have a term that we refer to as Place Value.

Place Value in mathematics can be defined as the value that a digit has based on its position or place in a number.

Examples of place value is:

Thousands represented by Th

Hundreds represented by H

Tens represented by T

Units represented by U

In the above question,

500 ÷ 50 gives us 10.

This is true because, 500 as a number, the Digit 5 has a place value of hundreds

50 as a number , the digit 5 has a place values of Tens.

10 as a number, the digit 1 has a place value of Tens.

In mathematics, the multiplication of 2 numbers in a place value of Tens always gives us a place value of hundreds.

For example, 10 × 50 = 500

Likewise, when we divide, a number with a place value of hundreds by a number with a place value of tens, we have a number with a place value of tens.

For example: 500 ÷ 50 = 10.

Therefore, the statement "500 divided by 50 is 10" is a true statement

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ArbitrLikvidat [17]

Answer:

Part 1) Option A. f of x equals one half times x plus 16

Part 2) Option A. x = 5

Part 3) Option C. x − 4y = −9

Part 4) Option A. The number of rides is the independent variable, and the total cost is the dependent variable.

Step-by-step explanation:

Part 1)

Let

x -----> the number of months

y ----> vertical jump in inches

step 1

Find the slope

we have the points

(0,16) and (2,17)

$m=(17-16)/(2-0)=\frac{1}{2}$

The equation of the line in slope intercept form is

$y=mx+b$

we have

$m=\frac{1}{2}$

$b=16$ -----> the point (0,16) is the y-intercept

substitute

$y=\frac{1}{2}x+16$

convert to function notation

f(x)=y

$f(x)=\frac{1}{2}x+16$

Part 2) What is the equation of a line that contains the points (5, 0) and (5, −2)?

step 1

Find the slope

we have the points

(5, 0) and (5, −2)

$m=(-2-0)/(5-5)=\frac{-2}{0}$

the slope is undefined

This is a vertical line (parallel to the y-axis)

therefore

The equation is

x=5

Part 3) Choose the equation that represents a line that passes through points (−1, 2) and (3, 1)

step 1

Find the slope

we have the points

(−1, 2) and (3, 1)

$m=(1-2)/(3+1)=-\frac{1}{4}$

step 2

Find the equation of the line into point slope form

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{4}$

$point\ (3, 1)$

substitute

$y-1=-\frac{1}{4}(x-5)$

Convert to standard form

Multiply by 4 both sides to remove the fraction

$4y-4=-(x-5)$

$4y-4=-x+5$

$x-4y=-4-5$

$x-4y=-9$

Part 4) Jewels has $6.75 to ride the ferry around Connecticut. It will cost her $0.45 every time she rides. Identify the dependent variable and independent variable in this scenario

we know that

The <u>independent variable</u> is the variable whose change isn’t affected by any other variable (An example the age and time)

The <u>dependent variable</u> it’s what changes as a result of the changes to the independent variable (An example of a dependent variable is how tall you are at different ages. The dependent variable (height) depends on the independent variable (age))

Let

x ----> the number of rides

y ----> the total cost in dollars

In this problem

The independent variable or input is the number of rides

The dependent variable or output is the total cost

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1 year ago

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2 years ago

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lord [1]

I suppose

$H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}$

The vectors that span $H$ form a basis for $P_2$ if they are (1) linearly independent and (2) any vector in $P_2$ can be expressed as a linear combination of those vectors (i.e. they span $P_2$ ).

Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of $H$ is 3.

Span:

Write an arbitrary vector in $P_2$ as $ax^2+bx+c$ . Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

$k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c$

which is equivalent to the system

$\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}$

so the vectors do span $P_2$ .

The vectors comprising $H$ form a basis for it because they are linearly independent.

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