<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B5x%2B3" id="TexFormula1" title="x^{2} +5x+3" alt="x^{2} +5x+3" align="absmidd

All categories

VikaD [51]

3 years ago

le" class="latex-formula">+2x^{2}+10x-15=0

Mathematics

1 answer:

il63 [147K]3 years ago

3 0

<h3>x²+5x+3+2x²+10x15 =0</h3><h3>x²+2x²+5x+10x+3+15=0</h3><h3>3x²+15x+18=0</h3><h3>3(x²+5x+6) =0 because 3 is common factor</h3><h3>3(x²+3x+2x+6) spill the middle term</h3><h3>3(x(x+3)+2(x+3) take the common factor from term</h3><h3>3(x+2) (x+3)</h3>

<h3>answer is 3(x+2) (x+3)</h3>

please mark this answer as brainlist

You might be interested in

Solve.<br>22 + 6-1<br>Your answer

Nimfa-mama [501]

Answer:

Step-by-step explanation:

22 + 6 = 28 - 1 = 27

6 0

3 years ago

Read 2 more answers

PLZ NEED HELP ASAP SHOW WORK TO PLZ IT WOULD MEAN A LOT

Marizza181 [45]

First Chart: Perimeter

Square Portion:

Original Side Lengths: P = 4 (1 + 1 + 1 + 1 ) =4

Double Side Lengths: P = 8 (2 x 4 = 8)

Triple Side Lengths: P = 12 (4 x 3 = 12)

Quadruple Side Lengths: P = 16 (4 x 4 = 16)

Rectangle Portion:

Original Side Lengths: P = 6 (1 x 2 + 2 x 2 = 6)

Double Side Lengths: P = 12 (2 x 2 + 4 x 2 = 12)

Triple Side Lengths: P = 24 (4 x 2 + 8 x 2 = 24)

Quadruple Side Lengths: P = 48 (8 x 2 + 16 x 2 = 48)

Second Chart: Area

Square Portion:

Original Side Lengths: A = 1 (1 x 1 = 1)

Double Side Lengths: A = 4 (2 x 2 = 4)

Triple Side Lengths: A = 9 (3 x 3 = 9

Quadruple Side Lengths: A = 16 ( 4 x 4 = 16)

Rectangle Portion:

Original Side Lengths: A = 2 ( 1 x 2 = 2 )

Double Side Lengths: A = 8 ( 2 x 4 = 8)

Triple Side Lengths: A = 18 ( 3 x 6 = 18)

Quadruple Side Lengths: A = 32 (4 x 8 = 32)

7 0

2 years ago

A right-angled triangle has shorter side lengths exactly c^2-b^2 and 2bc units respectively, where b and c are positive real num

yKpoI14uk [10]

Answer: hypotenuse = $c^{2} + b^{2}$

Step-by-step explanation: Pythagorean theorem states that square of hypotenuse (h) equals the sum of squares of each side ( $s_{1},s_{2}$ ) of the right triangle, .i.e.:

$h^{2} = s_{1}^{2} + s_{2}^{2}$

In this question:

$s_{1}$ = $c^{2}-b^{2}$

$s_{2} =$ 2bc

Substituing and taking square root to find hypotenuse:

$h=\sqrt{(c^{2}-b^{2})^{2}+(2bc)^{2}}$

Calculating:

$h=\sqrt{c^{4}+b^{4}-2b^{2}c^{2}+(4b^{2}c^{2})}$

$h=\sqrt{c^{4}+b^{4}+2b^{2}c^{2}}$

$c^{4}+b^{4}+2b^{2}c^{2}$ = $(c^{2}+b^{2})^{2}$ , then:

$h=\sqrt{(c^{2}+b^{2})^{2}}$

$h=(c^{2}+b^{2})$

Hypotenuse for the right-angled triangle is $h=(c^{2}+b^{2})$ units

3 0

3 years ago

Four friends are playing a game. They randomly choose a handful of marbles from a bag. The player with the highest ratio of blue

Monica [59]

We are given the number of Blue and Purple Marbles for each of the four friends. So first we have to find the ratio of blue marbles to purple marbles for each of them.

A) Rosa

Number of blue marbles = 10

Number of purple marbles = 12

Ratio of blue to purple marbles = $\frac{10}{12}=0.833$

B) Chi

Number of blue marbles = 5

Number of purple marbles = 6

Ratio of blue to purple marbles = $\frac{5}{6}=0.833$

C) Alberto

Number of blue marbles = 5

Number of purple marbles = 12

Ratio of blue to purple marbles = $\frac{5}{12}=0.417$

D) Pedro

Number of blue marbles = 10

Number of purple marbles = 6

Ratio of blue to purple marbles = $\frac{10}{6}=1.667$

From the above calculations we can see that Pedro has the highest value of ratio for blue marbles to purple marbles. Also we can see Pedro is the only one for whom the number of blue marbles is more than the number of purple marbles. So this confirms that our calculation is correct.

Thus, the player which gets to go first is Pedro.

4 0

3 years ago

Read 2 more answers

MARKING BRAINLY ASAP

lawyer [7]

Answer:

A is your answer

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers