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2 years ago

Solve each equation. check your sloution.7a=63

Mathematics

2 answers:

Marianna [84]2 years ago

7 0

Divide both sides by seven, 7 / 7 = 1, and 63 / 7 = 9, so:
a = 9

NNADVOKAT [17]2 years ago

5 0

63 divided by 7 is 9

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

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

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2 years ago

The cost of producing n necklaces is p(n) = 15n + 60. The necklaces cost $30, which can be represented by c(n) = 30n. For how ma

zysi [14]

Answer:

n=4

Step-by-step explanation:

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2 years ago

A spinner is divided into 10 equal parts and numbered from 1 through 10. What is the probability of spinning a number less than

konstantin123 [22]

Answer:

7/10 * 1/2 = 7/20

Step-by-step explanation:

$\frac{7}{10} * \frac{5}{10} \\\frac{7}{10} * \frac{1}{2} \\\frac{7}{20}$

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3 years ago

N = 3: – 3 and 4 + 2 i are zeros: f(- 1) = 58

Nimfa-mama [501]

Answer:

de eitcukqrncjakgkciwitjfuqjrjcjwjtjfkwjtkckskrkchakrjfnrjcjdnsnfncdn

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2 years ago

Solve for a 9a² = 16a College Intermediate Algebra

Fed [463]

We want to solve 9a² = 16a for a.

Because the a is on both sides, a good strategy is to get all the a terms on one side and set it equal to zero. Then we apply the Zero Product Property (if the product is zero then so are its pieces and Factoring.

9a² = 16a

9a² - 16a = 0 <-----subtract 16a from both sides

a (9a - 16) = 0 <-----factor the common a on the left side

a = 0 OR 9a - 16 =0 <----apply Zero Product Property

Since a = 0 is already solved we work on the other equation.

9a - 16 = 0

9a = 16 <----------- add 16 to both sides

a = 16/9 <----------- divide both sides by 9

Thus a = 0 or a = 16/9

7 0

3 years ago