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3 years ago

10

Can someone please help me factorize this one:

%7B1%7D%7B4%7D%20" id="TexFormula1" title="0.01y {}^{2} - \frac{1}{4} " alt="0.01y {}^{2} - \frac{1}{4} " align="absmiddle" class="latex-formula">

1 answer:

Scorpion4ik [409]3 years ago

4 0

First let us convert 0.01 into fraction:

$\frac{y^{2}}{100} -\frac{1}{4}$

Now we convert 100 and 4 into perfect squares as follows:

$(\frac{y}{10})^{2}-(\frac{1}{2})^{2}$

Now we use the property of :

$x^{2} -y^{2} =(x+y)(x-y)$

$(\frac{y}{10} +\frac{1}{2} ) (\frac{y}{10} -\frac{1}{2} )$

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1,000,000 divided by 6500 i will mark brainlest

Sveta_85 [38]

Answer:

153.846153846

Step-by-step explanation:

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3 years ago

Using distributive property combine 13c - 7c= Show work

emmasim [6.3K]

Actually, distributive property isn't needed to combine this. You can simply subtract 7 from 13 which will give you 6. The solution is 6c.

3 0

3 years ago

Read 2 more answers

Answer fast its sue by end of class

shtirl [24]

Step-by-step explanation:

question 1. Solve the equation 8-7/10 c = 6 - 1/5c for c

8 - 7/10c = 6 - 1/5c

subtract 8 from both sides:

8 - 7/10c - 8 = 6 - 1/5c - 8

- 7/10c = -2 - 1/5c

add 1/5c to both sides:

- 7/10c + 1/5c = -2 - 1/5c + 1/5c

- 7/10c + 1/5c = -2

change to common denominator:

- 7/10c+ 2/10c = -2

- 5/10c = -2

-1/2c = -2

multiply both sides by -2:

- 1/2c(-2) = -2(-2)

c = 4

___________________________________

question 2. 75 - 3.5y - 4y = 4y + 6 for y

75 - 3.5y - 4y = 4y + 6

75 - 7.5y = 4y + 6

add 7.5y to both sides:

75 - 7.5y + 7.5y = 4y + 6 + 7.5y

75 = 11.5y + 6

subtract 6 from both sides:

75 - 6 = 11.5y + 6 - 6

69 = 11.5y

divide both sides by 11.5:

69/11.5 = 11.5y/11.5

y = 6

___________________________________

Question 3. Solve the equation 16.5 + 2.75h = 9h + 7.5 − 4.25h for h.

16.5 + 2.75h = 9h + 7.5 − 4.25h

16.5 + 2.75h = 4.75h + 7.5

subtract 7.5 from both sides:

16.5 + 2.75h - 7.5 = 4.75h + 7.5 - 7.5

9 + 2.75h = 4.75h

subtract 2.75h from both sides:

9 + 2.75h - 2.75h = 4.75h - 2.75h

9 = 2h

divide both sides by 2:

9/2 = 2h/2

h = 9/2

6 0

1 year ago

The joint probability mass function of XX and YY is given by p(1,1)=0p(2,1)=0.1p(3,1)=0.05p(1,2)=0.05p(2,2)=0.3p(3,2)=0.1p(1,3)=

inessss [21]

Answer:

P(Y=1|X=3)=0.125

Step-by-step explanation:

Given :

p(1,1)=0

p(2,1)=0.1

p(3,1)=0.05

p(1,2)=0.05

p(2,2)=0.3

p(3,2)=0.1

p(1,3)=0.05

p(2,3)=0.1

p(3,3)=0.25

Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)

P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)

P(X=3)=p(3,1) +p(3,2) +p(3,3)

P(X=3)=0.05+0.1+0.25=0.4

$P(Y=1|X=3)=\frac{P(X=3,Y=1)}{P(X=3)} =\frac{p(3,1)}{P(X=3)}=\frac{0.05}{0.4}= 0.125$

Hence P(Y=1|X=3)=0.125

8 0

3 years ago

Which point would be a solution to the system of linear inequalities ?

brilliants [131]

Answer:

(12,-6)

Step-by-step explanation:

we have

$y\leq \frac{4}{3}x+5$ ----> inequality A

$y\geq -\frac{5}{2}x+5$ ---> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

<u><em>Verify each point</em></u>

Substitute the value of x and the value of y of each ordered pair in the inequality A and in the inequality B

case 1) (0,-1)

Inequality A

$-1\leq \frac{4}{3}(0)+5$

$-1\leq5$ ----> is true

Inequality B

$-1\geq -\frac{5}{2}(0)+5$

$-1\geq 5$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 2) (0,3)

Inequality A

$3\leq \frac{4}{3}(0)+5$

$3\leq5$ ----> is true

Inequality B

$3\geq -\frac{5}{2}(0)+5$

$3\geq 5$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 3) (-6,-6)

Inequality A

$-6\leq \frac{4}{3}(-6)+5$

$-6\leq-3$ ----> is true

Inequality B

$-6\geq -\frac{5}{2}(-6)+5$

$-6\geq 20$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 4) (12,-6)

Inequality A

$-6\leq \frac{4}{3}(12)+5$

$-6\leq21$ ----> is true

Inequality B

$-6\geq -\frac{5}{2}(12)+5$

$-6\geq -25$ ----> is true

therefore

The ordered pair is a solution of the system (makes true both inequalities)

8 0

3 years ago