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3 years ago

11

Andre placed some quarters and 1 dime into a money-counting machine. The machine reported that Andre has more than $5.00. Which

inequality can be solved to find the number of quarters, x, Andre has?

1 answer:

mario62 [17]3 years ago

5 0

The situation can be expressed by the inequality: 5 < 25x + 10

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215 divided by 32 with remainder

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quotient is 6 and the remainder is 23

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3 years ago

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Find the equation of the line which passes through the point (-4,12) and is perpendicular to the given line. Express your answer

Vika [28.1K]

Answer:

y = 1/7x + 88/7

Step-by-step explanation:

y = mx +b

since it is perpendicular you have to get the negative reciprocal of the slope which is 1/7

y = 1/7x + b then you would plug in your points

12 = 1/7(-4) + b and then solve for b

12 = -4/7 + b

+ 4/7

88/7 = b

y = 1/7x + 88/7

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Answer:

6 2/7

Step-by-step explanation:

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3 years ago

Jessie estimates the weight of her cat to be 10 pounds. The actual weight of the cat is 13.75 pounds. find the percent error.

romanna [79]

Answer:

$\large \boxed{27 \, \%} }$

Step-by-step explanation:

$\begin{array}{rcl}\text{Percent error}&= &\dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \,\%\\\\& = & \dfrac{\lvert 10 - 13.75\lvert}{13.75} \times 100 \, \% \\\\& = & \dfrac{\lvert -3.75\lvert}{13.75} \times 100 \, \%\\ \\& = & \dfrac{3.75}{13.75} \times 100 \,\%\\\\& = & 0.27 \times 100 \, \%\\& = & \mathbf{27 \, \%}\\\end{array}\\\text{The percent error is $\large \boxed{\mathbf{27 \, \%} }$}$

3 0

3 years ago

Rod is saving Php2000 in a bank at the end of each month which gives an interest of 1% compounded monthly.How much is the saving

agasfer [191]

Answer:

Php2040.38

Step-by-step explanation:

Given

$P = 2000$ --- Principal

$r = 1\%$ --- Rate

$t = 2\ years$ --- Time

$n = 12$ --- monthly

Required

Determine the amount at the end of two years

This is calculated as:

$A = P(1 + \frac{r}{n})^{nt}$

So, we have:

$A = 2000(1 + \frac{1\%}{12})^{12*2}$

$A = 2000(1 + \frac{1}{100*12})^{24}$

$A = 2000(1 + \frac{1}{1200})^{24}$

$A = 2000(\frac{1200+1}{1200})^{24}$

$A = 2000(\frac{1201}{1200})^{24}$

$A = 2000*1.02019$

$A = 2040.38$

<em>Hence, the final amount is: Php2040.38</em>

7 0

3 years ago