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Sonbull [250]
3 years ago
6

find 3 positive consecutive integers such that the product of the first and the third integer is 17 more than three times the se

cond integer

Mathematics
1 answer:
MariettaO [177]3 years ago
4 0

The required three consecutive numbers are 5,6 and 7.

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Point M is between points N and O on NO Find the length of NM if MO = 12.3 and NO = 26.9.
qaws [65]
NM + MO = NO
NM + 12.3 = 26.9
NM = 26.9 - 12.3
NM = 14.6

3 0
3 years ago
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Which expression correctly represents “ten times the difference of a number and six”?
krok68 [10]

Answer: 10(x-6)

Step-by-step explanation:

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3 years ago
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Please, help me on this question, guys! I know the answer for (a) so I'll write down, too. For (b) I tried "$89,000 because that
Alexeev081 [22]
Let x be a random variable representing the price of a Congo-imported black diamond. Let the higher price be p. Then,
P(x < p) = P(x < (p - mean)/sd) = P(x < (p - 60,430)/21,958.08) = P(z < 2)
Therefore,
(p - 60,430)/21,958.08 = 2
p - 60,430 = 2 x 21,958.08 = 43,916.16
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7 0
3 years ago
Any help would be appreciated. Thank you!
trapecia [35]

Answer:

Area =62.5\sqrt{6} square units

AB=5\sqrt{15} units

BC=5\sqrt{10} units

Step-by-step explanation:

In a right triangle the altitude drawn to the hypotenuse is the geometric mean of the segments at which this altitude divides the hypotenuse.

So,

BD^2=15\cdot 10\\ \\BD^2=150\\ \\BD=\sqrt{150}=5\sqrt{6}\ units

a. The area of the triangle ABC is

A_{ABC}=\dfrac{1}{2}\cdot BD\cdot AC=\dfrac{1}{2}\cdot 5\sqrt{6}\cdot (15+10)=\dfrac{125\sqrt{6}}{2}=62.5\sqrt{6}\ un^2.

b. The legs of the right triangle are geometric means of the segment adjacent to this leg and the hypotenuse, so

AB^2=AD\cdot AC=15\cdot 25\Rightarrow AB=5\sqrt{15}\ units\\ \\BC^2=CD\cdot AC=10\cdot 25\Rightarrow BC=5\sqrt{10}\ units

5 0
3 years ago
HALP MA PLS!!!!!!!!!!!!!!!!!!!!!!
Inessa05 [86]

Answer:

48 dollars.

Step-by-step explanation:

If you look up "40 percent off of 85" you'll get 48.

5 0
3 years ago
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