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Mashutka [201]
3 years ago
8

The sum of three consecutive odd integers is 75. find the value of all three integers.

Mathematics
2 answers:
sveta [45]3 years ago
7 0
23+25+27=75


Let me know if you need a detailed explanation.
SIZIF [17.4K]3 years ago
7 0
ANSWER: 23+25+27=75

Originally I had 24+25+26, but that's not odd consecutive integers. 
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Reduce in standard form -3/-15​
andriy [413]

Answer:

1/5? 0.2?

Step-by-step explanation:

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2 years ago
You purchased $132.49 worth of wheels and bearings for your skateboards. The shop charges $15 per board to install them. The tot
Mila [183]

4 skateborards will be repaired

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3 years ago
Help me the images are posted
Sav [38]

Answer:

28.

a. false

b. true

c. false

d. false

2. 8.4 X 2.3 X 2.09 = 40.3788

3. $10.04 - $13.47

4. Your friend would be incorrect because their decimal is in the wrong place. Take the decimals out of it and think about it. Can 1 times 2 equal anything near 31? no, so it's not reasonable that your friend could multiply 1.2 X 2.6 and get anything near 31.2

5. It is more reasonable to say between 5 and 6 because even when you round both of the numbers up (3 X 2), you still only get 6.

Step-by-step explanation:

ill put it in the comments so I can get this to you faster.

4 0
3 years ago
From a group of 12 students, we want to select a random sample of 4 students to serve on a university committee. How many combin
borishaifa [10]

Answer:

495 combinations of 4 students can be selected.

Step-by-step explanation:

The order of the students in the sample is not important. So we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

How many combination of random samples of 4 students can be selected?

4 from a set of 12. So

C_{n,x} = \frac{12!}{4!(8)!} = 495

495 combinations of 4 students can be selected.

8 0
3 years ago
I have nooooo clue, please help
Aleksandr-060686 [28]

Answer:

case a) x^{2}=3y ----> open up

case b) x^{2}=-10y ----> open down

case c) y^{2}=-2x ----> open left

case d) y^{2}=6x ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

y=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

x=a(y-k)^{2}+h

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

x^{2}=3y

so

y=(1/3)x^{2}

a=(1/3)

so

a>0

therefore

The parabola open up

case b) we have

x^{2}=-10y

so

y=-(1/10)x^{2}

a=-(1/10)

a

therefore

The parabola open down

case c) we have

y^{2}=-2x

so

x=-(1/2)y^{2}

a=-(1/2)

a

therefore

The parabola open to the left

case d) we have

y^{2}=6x

so

x=(1/6)y^{2}

a=(1/6)

a>0

therefore

The parabola open to the right

3 0
2 years ago
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