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Nuetrik [128]
2 years ago
11

ge\bold\red{{HELP}}" align="absmiddle" class="latex-formula">​

Mathematics
1 answer:
Serggg [28]2 years ago
5 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

A. <u>Understand</u>

The question is asking how much money is saved.

Given :

  • original price = 18795.50

  • lowered price = 15250.95

B. <u>Plan</u>

  • operation to be used here is subtraction

that is :

  • original price - lowered price

C. <u>Solve</u>

  • 18795.50 - 15250.95

  • 3544.55

Hence, savings = Php 3544.55

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4+ |2-3x| =7<br> solve for all values of x in simplest form
Simora [160]

Answer:

x = 1/12

1/12 = 0.833

Step-by-step explanation:

to find the value of x

4+ |2-3x| =7

8 - 12x = 7

-12x = 7-8

-12x = -1

x = 1/12

if we write 1/12 in simplest form so

1/12 = 0.833

5 0
3 years ago
What is the vaule of 0.77
padilas [110]
<span>seventy-seven hundredths.</span>
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Moira solved a problem on the board. What error did Moira make and how can she correct it?
iren [92.7K]

Answer:

you need to tell us what error she made first then you would find another way to solve the problem and look at all the posiibiltys

Step-by-step explanation:

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3 years ago
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Answer:

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8 0
3 years ago
HELP ME!!!
larisa [96]

Answer:

  -2, 8/3

Step-by-step explanation:

You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...

  A = (1/2)(f(a) +f(4))(4 -a)

  = (1/2)((3a -1) +(3·4 -1))(4 -a)

  = (1/2)(3a +10)(4 -a)

We want this area to be 12, so we can substitute that value for A and solve for "a".

  12 = (1/2)(3a +10)(4 -a)

 24 = (3a +10)(4 -a) = -3a² +2a +40

  3a² -2a -16 = 0 . . . . . . subtract the right side

  (3a -8)(a +2) = 0 . . . . . factor

Values of "a" that make these factors zero are ...

  a = 8/3, a = -2

The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.

_____

<em>Alternate solution</em>

The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.

5 0
3 years ago
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