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3 years ago

The unit rate for 5.99 for 13 Ib

Mathematics

2 answers:

Anit [1.1K]3 years ago

6 0

Answer:

0.46

Step-by-step explanation:

satela [25.4K]3 years ago

6 0

Answer:

$0.46

Step-by-step explanation:

To find the unit rate you have to divide, so our equation is:

5.99 ÷ 13 = 0.46 (rounded)

So the unit rate of 5.99:13 is 0.46:1

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HELPPPPPPPPPPPPPPPPPPPPPPPPP

Over [174]

Answer:

1: 11b(exponent of 2) +6b -9c

2: -11xy -9x -5x(exponent of 2)y

Step-by-step explanation:

8 0

3 years ago

Line l contains the points (1,5) and (4, -4). point P has the coordinates (-1, 1)

Colt1911 [192]

Answer:

The distance from p to l is $\sqrt{10}\ units$

Step-by-step explanation:

we know that

The distance between point p from line l is equal to the perpendicular segment from line l to point p

step 1

<em>Find the slope of line l</em>

we have the points

(1,5) and (4, -4)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{-4-5}{4-1}$

$m=\frac{-9}{3}$

$m=-3$

step 2

Find the equation of the line l

The equation in point slope form is equal to

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

we have

$m=-3$

$point\ (1,5)$

substitute

$y-5=-3(x-1)$ -----> equation A

step 3

Find the slope of the line perpendicular to the line l

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (The product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-3$ ---> slope of line l

therefore

$m_2=\frac{1}{3}$ ----> slope of the line perpendicular to line l

step 4

Find the equation of the line perpendicular to line l that passes through the point p

The equation in point slope form is equal to

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

we have

$m=\frac{1}{3}$

$point\ p(-1,1)$

substitute

$y-1=\frac{1}{3}(x+1)$ -----> equation B

step 5

Solve the system of equations

$y-5=-3(x-1)$ -----> equation A

$y-1=\frac{1}{3}(x+1)$ -----> equation B

Solve the system by graphing

The solution of the system is the intersection point both graphs

The solution is the point q(2,2)

see the attached figure

step 6

we know that

The distance between the point p and the line l is equal to the distance between the point p and the point q

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have the points

p(-1,1) and q(2,2)

substitute the values

$d_p_q=\sqrt{(2-1)^{2}+(2+1)^{2}}$

$d_p_q=\sqrt{(1)^{2}+(3)^{2}}$

$d_p_q=\sqrt{10}\ units$

3 0

3 years ago

soldi70 [24.7K]

we know that having similar triangles so the ratio of sides,area and volume are same.

$s/s=s^2/s^2$

we know that $s^2$ will be the area of triangle.

$s^2/s^2=75/12$

so $s/s=\sqrt{75/12}$

so the ratio of the perimeter of two triangles will be

$(s+s+s)/(s+s+s)=(\sqrt{75} +\sqrt{75} +\sqrt{75} )/(\sqrt{12}+\sqrt{12} +\sqrt{12} )$

ratio of the perimeter= $\sqrt{225} /\sqrt{36}$

8 0

3 years ago

Read 2 more answers

Which of the following employees derives their income from a fee for the items sold?

dezoksy [38]

Answer:

commission

Step-by-step explanation:

3 0

3 years ago

Suppose that $9500 is placed in an account that pays 8% interest compounded each year.

Blababa [14]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$9500\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{each year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 1}\implies A=9500(1.08)\implies \boxed{A=10260} \\\\\\ \stackrel{\textit{after 2 years, t = 2}}{A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 2}}\implies A=9500(1.08)^2\implies \boxed{A=11080.8}$

6 0

3 years ago