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4 years ago

4x + 2y = –22 and 4x + 4y = -12

Mathematics

2 answers:

mart [117]4 years ago

5 0

hope it helps

plz mark brainliest

Artyom0805 [142]4 years ago

4 0

So this is how we solve this question by cancelling x value in both equations for obtaining y and then substituting y in equation to get x

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What number is39% of 120

noname [10]

46.8 is 39% of 120. Hope this helped you!

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3 years ago

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To the brainly comunity TYSM FOR HELPIN PEOPLE!!!!!

umka21 [38]

Answer:

i need BRAINLY plz

Step-by-step explanation:

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3 years ago

Stefan's rectangular bedroom is 12 feet by 9 feet. What is the diagonal distance from one

Rainbow [258]

Answer:

505 ft

Step-by-step explanation:

pythagorean therom = $a^{2} +b^{2}=c^{2}$

a= 15

b=9

$12^{2} +19^{2}=c^{2}$

144+361=505

505 ft

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3 years ago

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Buddy, Louis, Mabel and Spot are all puppies and are playing in the forest. They notice seven bees and five cockroaches, and the

Natasha2012 [34]

Answer:

188 legs

Step-by-step explanation:

A game of legs!!

In this question, we are asked to calculate the number of legs of animals present in the question.

Let’s first identify the individual brands of animals present. We have ;

4 dogs, 7 bees, 5 cockroaches, 8 spiders, 3 deers, 4 boars and 2 antlers.

A dog has 4 legs; this means total number of legs contributed by the dogs to the pool will be 4 * 4 = 16 legs

A bee has six legs; total number of bees legs is thus 7 * 6 = 42 legs

A cockroach has 6 legs; total number of cockroaches legs is 6 * 5 = 30 legs

8 spiders have a total of 8 * 8 = 64 legs

3 deers, 4 boars and 2 antlers have 4 legs each making a total of 9 * 4 = 36 legs

The total number of legs present is thus; 36 + 64 + 30 + 42 + 16 = 188 legs

6 0

3 years ago

Find the tenth term of the expansion (x+ y)¹³

Anon25 [30]

Answer:

$715x^4y^9$

Step-by-step explanation:

Given

$(x + y)^{13$

Required

Determine the 10th term

Using binomial expansion, we have:

$(a + b)^n = ^nC_0a^nb^0 + ^nC_1a^{n-1}b^1 + ^nC_2a^{n-2}b^2 +.....+^nC_na^{0}b^n$

For, the 10th term. n = 9

So, we have:

$(a + b)^n = ^nC_{9}a^{n-9}b^{9$

$(x + y)^{13} = ^{13}C_{9}x^{13-9}y^9$

$(x + y)^{13} = ^{13}C_{9}x^4y^9$

Apply combination formula

$(x + y)^{13} = \frac{13!}{(13-9)!9!}x^4y^9$

$(x + y)^{13} = \frac{13!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10*9!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4*3*2*1}x^4y^9$

$(x + y)^{13} = \frac{17160}{24}x^4y^9$

$(x + y)^{13} = 715x^4y^9$

Hence, the 10th term is $715x^4y^9$

3 0

3 years ago