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2 years ago

12

Susanna can work 10 math problems in 15 minutes. How many can she work in an hour and a half?

2 answers:

shusha [124]2 years ago

7 0

Well this one is easy!
10 ~ 15
20 ~ 30
30 ~ 45
40 ~ 60
50 ~ 75
60 ~ 90
90 = 1 hour and 30 minutes! < 60 math problems.

docker41 [41]2 years ago

4 0

She can work 60 in an hour and a half. We find this by figuring out how many minutes are in an hour and a half (there are 90 minutes), dividing that number by 15 (which equals 6), and multiplying that number by 10 (which equals 60). 60 problems is the answer.

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Answer:

(a) Neither

(a) Perpemdicular

Step-by-step explanation:

Required

Determine the relationship between given lines

(a)

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and

$y = \frac{2}{5}x + 7$

An equation written in form: $y=mx + b$ has the slope:

$m \to slope$

So, in both equations:

$m_1 = -\frac{2}{5}$

$m_2 = \frac{2}{5}$

For both lines to be parallel

$m_1 = m_2$

This is false in this case, because:

$-\frac{2}{5} \ne \frac{2}{5}$

For both lines to be perpendicular

$m_1 * m_2 = -1$

This is false in this case, because:

$-\frac{2}{5} * \frac{2}{5} \ne -1$

(b)

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and

$-5x + 2y = 4$

Write equations in form: $y=mx + b$

$5y + 2x = -10$

$5y = -2x +10$

Divide by 5

$y = -\frac{2}{5}x +2$

$-5x + 2y = 4$

$2y = 5x + 4$

Divide by 2

$y = \frac{5}{2}x + 2$

In both equations:

$m_1 = -\frac{2}{5}$

$m_2 = \frac{5}{2}$

For both lines to be parallel

$m_1 = m_2$

This is false in this case, because:

$-\frac{2}{5} \ne \frac{5}{2}$

For both lines to be perpendicular

$m_1 * m_2 = -1$

This is true in this case, because:

$-\frac{2}{5} * \frac{5}{2} = -1$

Cancel out 2 and 5

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-by-step explanation:

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