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3 years ago

Check the iinstructions

Mathematics

1 answer:

pishuonlain [190]3 years ago

5 0

Answer:

Cost of small box of oranges = 7

Cost of small box of oranges = 13

Step-by-step explanation:

Step 1) Let the cost of small box of oranges = x

Let the cost of small box of oranges = y

Step 2)

Equation 1) Matt sells 3 small boxes and 14 large boxes for Php. 203

3x + 14y = 203 -------------------(I)

Equation 2) Ming sells 11 small boxes and 11 large boxes for Php. 220

11x + 11y = 220 ----------------(II)

Step 3:

Multiply equation (I) by 11 and equation (II) by (-3).

(I)*11 33x + 154y = 2233

(II)*(-3) -33x - 33y = -660 {Now add and x will be eliminated}

121y = 1573

y = 1573/121

y = 13

Substitute y = 13 in equation (II)

11x + 11*13 = 220

11x + 143 = 220

11x = 220 - 143

11x = 77

x = 77/11

x = 7

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3 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 $8y^{6}+ 144y^{5}+ 640y^{4}$

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

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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3 years ago

How to solve this question

Viefleur [7K]

Answer:

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Step-by-step explanation:

7(x+3)=6-(-7x - 15)

L.H.S

=7(x+3)

=7x+21 (multiply 7 by (x+3))

R.H.S:

=6-(-7x - 15)

= 6+7x+15 (multiply the -ive sign in the bracket)

= 7x + 21 ( adding 15 and 6)

now compare the two sides

7x+21 = 7x + 21 hence we prove that L.H.S= R.H.S

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