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3 years ago

Answer correctly !!!!!! Will mark brainliest !!!!!!!!

Mathematics

1 answer:

morpeh [17]3 years ago

5 0

Answer:

(x+9)(x+2)

Step-by-step explanation:

x(x+9)+2(x+9)

(x+2)(x+9)

the order of the (---) doesn't matter

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What is the value of x? 3(2X-5)-4x=11

kupik [55]

Answer:

x = 13

Explanation:

3(2x−5)−4x=11

Step 1: Simplify both sides of the equation

3(2x−5)−4x=11

(3)(2x)+(3)(−5)+−4x=11 (Distribute)

6x+−15+−4x=11

(6x+−4x)+(−15)=11 (Combine Like Terms)

2x+−15=11

2x−15=11

Step 2: Add 15 to both sides

2x−15+15=11+15

2x=26

Step 3: Divide both sides by 2

2x / 2 = 26 / 2

x = 13

3 0

3 years ago

Read 2 more answers

Please help me solve for the value of x step by step.

swat32

Ok, so here's how you do it.Since you have those two 21s, you know those sides are equal, which means that you're gonna make an equation 38+2x=180 because all angles in a triangle have to equal 180 and whatever that x equal is also what that other missing angle is I think because of those two sides being equal and then you're going to solve that equation for x, so you wanna get x by itself. You subtract that 38 from the sides it's on and do the same to the other side, so 180-38, which equals 142. So now you have 2x=142, you have to get x by itself, so to do that you divide by 2 on both sides, so 142/2=71, so x=71.Hope that helps.

7 0

3 years ago

If a couple were planning to have three children, the sample space summarizing the gender outcomes would be: bbb, bbg, bgb,

harkovskaia [24]

Answer:

a.S={hh,sh,hs,ss}

b.tex]\frac{1}{4}[/tex]

c. $\frac{1}{2}$

Step-by-step explanation:

We are given that a sample space of three children

S={bbb,bbg,bgb,bgg,gbb,gbg,ggb,ggg}

a.We have to construct similar space for two children where h for healthy and s for sick.

Then the sample space of two children

S={hh,sh,hs,ss}

b.Number of cases favorable to two healthy children=1

Total number of cases=4

Number of cases for two healthy children=1

Probability =Number of favorable cases divided by total number of cases

Probability= $\frac{1}{4}$

Hence, the probability of getting two healthy children= $\frac{1}{4}$

c.We have to find the probability of getting exactly one healthy child and one sick child

Number of cases of one healthy child and one sick child={hs,sh}=2

Probability= $\frac{2}{4}=\frac{1}{2}$

Hence, the probability of getting exactly one healthy child and one sick child= $\frac{1}{2}$

6 0

3 years ago

Jane wrote all natural numbers from 5 to 12. including 5 and 12.

Elina [12.6K]

Answer:

5, 6, 7, 8, 9, 10, 11 and 12 are the natural numbers between 5 and 12 including them, therefore Jane wrote 8 numbers. natural numbers start at positive 1 and go up by whole numbers infinitely, hope this helps

8 0

3 years ago

Cos^2 x+4sin^2 x/2=1

lana [24]

$Let\ \dfrac{x}{2}=a,\ therefore\ x=2a.\\\\\cos^2x+4\sin^2\dfrac{x}{2}=\cos^22a+4\sin^2a\\\\\text{use}\ \cos2x=\sin^2x-\cos^2x\\\\=(\sin^2a-\cos^2a)^2+4\sin^2a\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(\sin^2a)^2-2(\sin^2a)(\cos^2a)+(\cos^2a)^2+4\sin^2a\\\\=\sin^4a-2\sin^2a\cos^2a+\cos^4a+4\sin^2a\\\\=\underbrace{\sin^4a+2\sin^2a\cos^2a+\cos^4a}_{(*)}-4\sin^2a\cos^2a+4\sin^2a\\\\\text{use}\ (*)\qquad(a+b)^2=a^2+2ab+b^2$

$=\underbrace{(\sin^2a)^2+2\sin^2a\cos^2a+(\cos^2a)^2}_{(*)}-4\sin^2a(\cos^2a-1)\\\\=(\sin^2a+\cos^2a)^2-4\sin^2a(\cos^2a-1)\\\\\text{use}\ \sin^2a+\cos^2a=1\to\sin^2a=\cos^2a-1\\\\=1^2-4\sin^2a(\sin^2a)=1-4\sin^4a=1-(2\sin^2a)^2$

$\cos^22a+4\sin^2a=1\\\\1-(2\sin^2a)^2=1\qquad\text{subtract 1 from both sides}\\\\-(2\sin^2a)^2=0\to2\sin^2a=0\qquad\text{divide both sides by 2}\\\\\sin^2a=0\to\sin a=0\\\\a=k\pi\ for\ k\in\mathbb{Z}\\\\\dfrac{x}{2}=k\pi\qquad\text{multiply both sides by 2}\\\\\boxed{x=2k\pi\ for\ k\in\mathbb{Z}}$

6 0

3 years ago