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2 years ago

A car is traveling at 25 miles per hour on a residential street. About how many feet does it travel per second?

Mathematics

1 answer:

Oksanka [162]2 years ago

8 0

Answer:

Abt 37

Step-by-step explanation:

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Answer this problem. Simplify a. 63.19 b. 25.84 c. 81.49 d. 94.57

just olya [345]

81.49 is the answer. Have a great day!

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3 years ago

Based on the information below, choose the correct answers.

lozanna [386]

Since there are 120 payment and the total interest paid was $ 14,644.95. so the average monthly interest can be calculated by:

Average monthly interest = $ 14,644.95 / 120

Average monthly interest = $ 122.04

% total interest = ($ 14,644.95) / $ 39,644.95 x 100

% total interest =36.94 %

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2 years ago

Stephen uses 2/25 kilograms of tofu in each serving of his famous tofu dish. He has 1 1/10 kilograms of tofu. How many servings

larisa [96]

Answer: He has 1 1/10 kilograms of tofu.

Step-by-step explanation: We will have to divide the number of tofu required with the number of tofu own. Stephen uses 2/25 kilograms of tofu in each serving and he has 11/10 kilograms tofu.

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2 years ago

The two points on the coordinate plane represent Jane's house and her friend's house. Find the distance between the houses. Ques

Nesterboy [21]

Answer:

if it helps then mark brainliest

8 0

2 years ago

Read 2 more answers

A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for $83 or a chair a

vesna_86 [32]

$Chair=x$

$Table=y$

$Desk=z$

$\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}$

keep the first row as normal, then in the other ones, we can isolate Y and X

$\begin{Bmatrix}x+y&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace at first row...

$\begin{Bmatrix}(77-z)+(83-z)&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}160-2z&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&40-160\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&-120\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}z&=&60\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace the Z to discovery the other value

$\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}$

$\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}$

6 0

3 years ago