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Nikolay [14]
3 years ago
12

Find five consecutive integers such that the sum of the first and 5 times the third is equal to 41 less than 3 times the sum of

the second fourth and fifth
Mathematics
1 answer:
bogdanovich [222]3 years ago
6 0

Answer:

see below

Step-by-step explanation:

We'll cal the first integer x and then the rest of them will be x + 1, x + 2, x + 3 and x + 4. We can write x + 5(x + 2) = 3(x + 1 + x + 3 + x + 4) - 41.

x + 5x + 10 = 3(3x + 8) - 41

6x + 10 = 9x + 24 - 41

6x + 10 = 9x - 17

3x = 27

x = 9

The numbers are 9, 10, 11, 12, 13.

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To evaluate the expression all you have to do is substitute the value for the variable.

3 + y + 6, y = 5
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14.

The solution is 14.
8 0
3 years ago
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What is 3/5 x (times/multiply by) 3/8?
Gennadij [26K]

Answer:

9/40

Step-by-step explanation:

3/5 * 3/8

Multiply the numerators

3*3 = 9

Multiply the denominators

5*8 = 40

9/40

The fraction does not simplify

5 0
2 years ago
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Solve x to the power of 2 = 4 where x is a real number. Please help fast
zimovet [89]

Answer:

It's 2

Step-by-step explanation:

X is 2 because if you multiply two by two you get 4

7 0
2 years ago
Hey guys, I need some help. Would appreciate it.
34kurt

the 3th choice because it does not add up to 180, it actually adds up to something greater

7 0
3 years ago
find the surface area of a right regular hexagonal pyramid with sides 3 cm and slant heights 6cm. show all of your work.​
DedPeter [7]

Answer:

The surface area of right regular hexagonal pyramid = 82.222 cm³

Step-by-step explanation:

Given as , for regular hexagonal pyramid :

The of base side = 3 cm

The slant heights = 6 cm

Now ,

The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times a^{2} + 3 a \sqrt{h^{2}+ 3\times \frac{a^{2}}{4}}

Where a is the base side

And h is the slant height

So, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 3^{2} + 3 \times 3 \sqrt{6^{2}+ 3\times \frac{3^{2}}{4}}

Or, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 9 + 9 \sqrt{36+ 3\times \frac{9}{4}}

Or,  The surface area of right regular hexagonal pyramid = 23.38 + 9 × \frac{171}{4}

∴  The surface area of right regular hexagonal pyramid = 23.38 + 9 × 6.538

I.e The surface area of right regular hexagonal pyramid = 23.38 + 58.842

So,  The surface area of right regular hexagonal pyramid = 82.222 cm³ Answer

6 0
3 years ago
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