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2 years ago

A store sells candy at $.50, $1, $1.50, $2, and $3 per kilogram. How many kilograms of each kind of candy does $3 buy?

Mathematics

1 answer:

evablogger [386]2 years ago

8 0

Money : price = quantity

1) 3 : 0.5 = 6 Kg
2) 3 : 1 = 3 Kg
3) 3 : 1.5 = 2 Kg
4) 3 : 2 = 1.5 Kg
5) 3 : 3 = 1 Kg

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A line passes through the points (–1, –5) and (4, 5). The point (a, 1) is also on the line.

IgorC [24]

Answer:

a = 2

Step-by-step explanation:

Let's find the equation of the line using the 2 points given.

Let's call -1 as x_1 and -5 as y_1

also, 4 as x_2 and 5 as y_2

The equation of line is :

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Let's plug the points and get the equation of the line:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\y+5=\frac{5+5}{4+1}(x+1)\\y+5=2(x+1)\\y+5=2x+2\\y=2x-3$

Now, to find a, we substitute a in x and 1 in y of the equation of the line we just got:

$y=2x-3\\1=2(a)-3\\1=2a-3\\2a=4\\a=\frac{4}{2}=2$

4 0

3 years ago

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−2y−8+4yminus, 2, y, minus, 8, plus, 4, y

lora16 [44]

Answer:

a number 12

Step-by-step explanation:

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2 years ago

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Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

3 years ago

Find the slope of the tangent line to the curve f(x)=e^(x) at (0.4,1.49)

almond37 [142]

Answer:

1.49

Step-by-step explanation:

In order to find the slope of the tangent line to a given equation, and in a given point, we need to:

1. Find the first derivative of the given function.

2. Evaluate the first derivative function in the given point.

1. Let's find the first derivative of the given function:

The original function is $f(x)=e^{x}$

But remeber that the derivative of $e^{x}$ is $e^{x}$

so, $f'(x)=e^{x}$

2. Let's evaluate the first derivative function in the given point

The given point is (0.4,1.49) so:

$f'(x)=e^{x}$

$f'(0.4)=e^{0.4}$

$f'(x)=1.49$

Notice that the calculated slope of the tangent line is equal to the y-coordinate of the given point because f'(x)=f(x). In conclusion, the slope of the tangent line is equal to 1.49.

8 0

3 years ago

Compare √7 and 1.9? > < = Help pls 

Paraphin [41]

Answer:

that answer is 7 is bigger then 1.9

Step-by-step explanation:

so 7 is a whole and 1.9 is a decimal

8 0

2 years ago

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