What is the answer to 4a+10b+12d. what is b

Mathematics

1 answer:

gregori [183]2 years ago

5 0

B is a variable that is as far as it can be solved

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Suppose the sum of two integers is negative. Must both integers be negative? Explain.

mash [69]

Answer:

Not necessarily. There are many ways to write a basic equation with a negative answer. For example, -3-4 = -7. Here 4 is a positive number but because you subtract 4 to 3 or take away 4 from 3 being a negative number and so you get a negative answer. Another example is 6+(-9). There are a couple of ways you can resolve this. My method is to subtract 9 from 6 which gives you 3 and simply add a negative sign.

Let me know if you'd like more examples. Hope this helps!! Sorry if it is confusing I can explain you in a more simpler way if you'd like.

Step-by-step explanation:

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2 years ago

Jamal ran 7 kilometers, sprinted 400 meters, and climbed bleachers for a distance of 1.6 kilometers during track practice. How m

seraphim [82]

First, convert 7 kilometers to meters, which is 7000 meters.
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4 0

3 years ago

Read 2 more answers

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