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2 years ago

-15+32-7-1 simplify this question

Mathematics

2 answers:

nikitadnepr [17]2 years ago

4 0

Answer:

Step-by-step explanation:

32-15 is 17. 17-7 is 10, and since we're missing the 1, 10-1 is 9. Therefore, the answer is 9.

Effectus [21]2 years ago

4 0

The answer is 9............

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Consider this function.

charle [14.2K]

Answer:

3. Since the range of the original function is limited to y> 6, the domain of the inverse function is x ≥ 6.

Step-by-step explanation:

The domain of a function is the range of its inverse, and vice versa. The only answer choice that expresses this relationship is choice 3.

Comment on the answer choice:

The slope of the function is undefined at x=4, so restricting the function domain to the portion with positive slope means the domain restriction of the function is x > 4. That also means the range restriction of the function is y > 6. The domain restriction of the inverse function is the same: x > 6, not x ≥ 6. The answer choice has an error.

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3 years ago

Read 2 more answers

Help Please!!!!!!!!!!!!!

zheka24 [161]

Answer:

A. 2086 yd²

Step-by-step explanation:

SA = 2lw + 2wh + 2lh
SA = 2*17*20 + 2*20*19 + 2*17*19
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SA = 2086
Hope this helps :)

3 0

2 years ago

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$