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3 years ago

11

A teacher working's day is 7 1/2 hours long. one certain day, he spent 3 1/2 hours teaching and 2 1/2 hours marking student assi

gnment. What fraction did he spend (a) Teaching (b) Marking.

1 answer:

SpyIntel [72]3 years ago

4 0

Answer:

(a) 7/15

(b) 1/3

Step-by-step explanation:

(a) teaching:

(3 1/2) / (7 1/2) = (7/2) / (15/2) = 7/2 * 2/15 = 7/15

(b) marking:

(2 1/2) / (7 1/2) = (5/2) / (15/2) = 5/2 * 2/15 = 5/15 = 1/3

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Please help I will mark brainliest!!!!!!!!!!

Oxana [17]

Answer:

9/10 pounds

Step-by-step explanation:

3/5 × 1 1/2

= 9/10

5 0

2 years ago

Cary is 7 years older than Dan. In 7 years the sum of their ages will be 79. Find the age of each man now.

Vilka [71]

Answer:

Dan is 29. Cary is 36.

Step-by-step explanation:

Let Dan's age now be D.

Let Cary's age now be C.

"Cary is 7 years older than Dan."

C = D + 7

"In 7 years the sum of their ages will be 79. "

In 7 years, the ages of Dan and Cary will be D + 7 and C + 7, respectively.

D + 7 + C + 7 = 79

C + D + 14 = 79

C + D = 65

C = D + 7

The last two equations in bold above form a system of equations. Since the second equation is already solved for C, we use the substitution method.

D + 7 + D = 65

2D + 7 = 65

2D = 58

D = 29

C = D + 7

C = 29 + 7

C = 36

Answer: Dan is 29. Cary is 36.

3 0

3 years ago

The exponential function f(x) = 3(5)x grows by a factor of 25 between x = 1 and x = 3. What factor does it grow by between x = 5

Delvig [45]

Answer:

B) 25

Step-by-step explanation:

Given exponential function:

$f(x)=3(5)^x$

The growth factor between $x=1$ and $x=3$ is 25.

To find the growth factor between $x=5$ and $x=7$

Solution:

The growth factor of an exponential function in the interval $x=a$ and $x=b$ is given by :

$G=\frac{f(b)}{f(a)}$

We can check this by plugging in the given points.

The growth factor between $x=1$ and $x=3$ would be calculated as:

$G=\frac{f(3)}{f(1)}$

$f(3)=3(5)^3$

$f(1)=3(5)^1$

Plugging in values.

$G=\frac{3(5)^3}{3(5)^1}$

$G=\frac{(5)^3}{(5)^1}$ (On canceling the common terms)

$G=(5)^{(3-1)}$ (Using quotient property of exponents $\frac{a^b}{a^c}=a^{(b-c)}$ )

$G=(5)^{2}$

∴ $G=25$

Similarly the growth factor between $x=5$ and $x=7$ would be:

$G=\frac{f(7)}{f(5)}$

$f(7)=3(5)^7$

$f(5)=3(5)^5$

Plugging in values.

$G=\frac{3(5)^7}{3(5)^5}$

$G=\frac{(5)^7}{(5)^5}$ (On canceling the common terms)

$G=(5)^{(7-5)}$ (Using quotient property of exponents $\frac{a^b}{a^c}=a^{(b-c)}$ )

$G=(5)^{2}$

∴ $G=25$

Thus, the growth factor remains the same which is =25.

3 0

2 years ago

Read 2 more answers

Identify the slope and y-intercept of the graph of the equation Y=3/4x-2

sukhopar [10]

Hey there! :)

y = 3/4x - 2

To find the slope and y-intercept, simply use slope-intercept form to understand this.

Slope-intercept form is : y = mx + b ; where m = slope, b = y-intercept.

Therefore, you must ask yourself : which value is in the "m" spot and which value is in the "b" spot in our given equation?

Using this logic, we can come to the conclusion that :

Slope = 3/4 & Y-intercept = -2

~Hope I helped! :)

6 0

3 years ago

Question 8 (1 point)

sergey [27]

Y=-5x+4 y=7x+5 that is the line

3 0

3 years ago