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Sedaia [141]
3 years ago
15

Write a real-world problem that could be represented by the expression 3(23). Then explain how you could solve the problem menta

lly.
Mathematics
1 answer:
Natalija [7]3 years ago
5 0
3(23) really means 3*23, or just 3 multiplied by 23.  For example: "Jana has 23 dollars.  Her friend John has $23, and his friend Jacob also has 23 dollars.  How much money do they have all together?"  To solve this problem mentally, you could break 3*23 into two numbers you're familiar with multiplying, like 20 and 3, and then add the numbers together.  For example: 3*20 + 3*3.  3*20=60, 3*3=9, and 60+9=69.  So, 3*23=69.
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PLZ HELP I WILL MARK BRAINLYIST!!!!!!!!!!!!!!!!Zoe is doing a survey to find the time her classmates spend playing video games.
Radda [10]

Answer:

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Step-by-step explanation:

3 0
3 years ago
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A pole is 3.4 m tall casts a shadow that is 1.8 m long. At the same time, a nearby tower casts a shadow that is 39.75 m long. Ho
KATRIN_1 [288]

Answer:

75

Step-by-step explanation:

For this, you need to solve a proportion-

3.4      x

----- = -----

1.8      39.75

3.4*39.75=135.15

135.15/1.8= 75.0833.....

75.0833.. is rounded to 75

7 0
3 years ago
Can someone please help me factor and write the equations for these two problems ?
prisoha [69]

Answer:

Step-by-step explanation:

Asymptotes 3

g(x) = \frac{3x}{x^2-3x-10}

Factors of denominator will be,

x² - 3x - 10 = x² - 5x + 2x - 10

                 = x(x - 5) + 2(x - 5)

                 = (x + 2)(x - 5)

Therefore, factored form of g(x) will be,

g(x) = \frac{3x}{(x + 2)(x - 5)}

Asymptotes 4

h(x) = \frac{(x-5)}{x^{2} + 14x + 40}

      = \frac{x-5}{x^{2}+10x+4x+40}

      = \frac{x-5}{x(x+10)+4(x+10)}

      = \frac{x-5}{(x+10)(x+4)}

5 0
3 years ago
What is the value of the missing angle. <br> A. 50<br> B. 60<br> C. 100<br> D. 240
34kurt
The answer is b 60 degrees <span />
5 0
3 years ago
Read 2 more answers
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

5 0
3 years ago
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