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3 years ago

11

Drag numbers to use a negative factor to factor the expression –5g + 15h – 25. Numbers may be used once, more than once, or not

at all.
–1 –2 –3 –4 –5 1 2 3 4 5
(_____g –_____ h +_____)

1 answer:

Gala2k [10]3 years ago

6 0

Answer:

(5g –15h +25)

hope dis helps ^-^

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There are 5 red balls, 4 blue balls, 6 yellow balls and 10 green balls in a box,

lys-0071 [83]

<h2 /><h2>Here we go ~ </h2>

According to given information there are :

5 red balls

4 blue balls

6 yellow balls

10 green balls

<h3>1. what is the probability that the ball chosen is red ?</h3>

-

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{total \: red \: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{5}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{5}{25}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{1}{5}$

<h3>2. what is the probability that the ball chosen is blue ?</h3>

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{total \: blue \: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{4}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{4}{25}$

<h3>3. what is the probability that the ball chosen is yellow ?</h3>

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{total \: yellow\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{6}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{6}{25}$

<h3>4. what is the probability that the ball chosen is green ?</h3>

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{total \: green\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{10}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{10}{25}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{2}{5}$

<h3>5. what is the probability that the ball chosen is not green ?</h3>

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{total \: non \: green\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{5 + 4 + 6}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{15}{25}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{3}{5}$

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What are the odds of rolling 2 dice and getting the sum of 7?

Yuliya22 [10]

Answer:

1/6

Step-by-step explanation:

favorable over total outcomes

favorable is 6

total is 36

6/36 = 1/6

6 favorable outcomes are:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

total of 36 outcomes are :

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

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Answer:

3: $a^{2}+2ab+b^{2}$

4: $8x^{3} -12x^{2}+6x-1$

Step-by-step explanation:

Question 3 can be re-written as:

$(a+b)(a+b)$

which equals:

$a^{2}+2ab+b^{2}$

Question 4 can be re-written as:

$(2x-1)(2x-1)(2x-1)$

Doing the first 2 brackets gives us:

$(4x^{2} -4x+1)(2x-1)$

Expanding these two:

$8x^{3} -4x^{2} -8x^{2} +4x+2x-1$

Simplifying it:

$8x^{3} -12x^{2}+6x-1$

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