All categories

2 years ago

Simplify: (3a – 4)2

Mathematics

2 answers:

Nadya [2.5K]2 years ago

7 0

Most likely sure this is correct 2 times (3a-4)

Mrac [35]2 years ago

5 0

(3a - 4)²
= (3a)² - 2*3a*4 + 4²
= 9a² - 24a + 16

You might be interested in

A bag contains 8 red balls and 4 white balls. Two balls are drawn without replacement. (Enter your probabilities as fractions.)

RoseWind [281]

Answer:

total marbles;12

after drawing red total counts are 7 red and 4 white 11 in total

probabilities are 4/11

6 0

2 years ago

COMPUTE 3 ( 2 1/2 - 1 ) + 3/10

Juli2301 [7.4K]

Answer:

<h3> $\boxed{ \frac{24}{5} }$ </h3>

Step-by-step explanation:

$\mathsf{3(2 \frac{1}{2} - 1) + \frac{3}{10} }$

Convert mixed number to improper fraction

$\mathrm{3( \frac{5}{2} - 1) + \frac{3}{10} }$

Calculate the difference

⇒ $\mathrm{3( \frac{5 \times 1}{2 \times 1} - \frac{1 \times 2}{1 \times 2} }) + \frac{3}{10}$

⇒ $\mathrm{ 3 \times( \frac{5}{2} - \frac{2}{2}) } + \frac{3}{10}$

⇒ $\mathrm{3 \times ( \frac{5 - 2}{2} ) + \frac{3}{10} }$

⇒ $\mathrm{3 \times \frac{3}{2} + \frac{3}{10} }$

Calculate the product

⇒ $\mathrm{ \frac{3 \times 3}{1 \times 2} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{9}{2} + \frac{3}{10}}$

Add the fractions

⇒ $\mathsf{ \frac{9 \times 5}{2 \times 5} + \frac{3 \times 1}{10 \times 1} }$

⇒ $\mathrm{ \frac{45}{10} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{45 + 3}{10 } }$

⇒ $\mathrm{ \frac{48}{10} }$

Reduce the numerator and denominator by 2

⇒ $\mathrm{ \frac{24}{5} }$

Further more explanation:

Addition and Subtraction of like fractions

While performing the addition and subtraction of like fractions, you just have to add or subtract the numerator respectively in which the denominator is retained same.

For example :

Add : $\mathsf{ \frac{1}{5} + \frac{3}{5} = \frac{1 + 3}{5} } = \frac{4}{5}$

Subtract : $\mathsf{ \frac{5}{7} - \frac{4}{7} = \frac{5 - 4}{7} = \frac{3}{7} }$

So, sum of like fractions : $\mathsf{ = \frac{sum \: of \: their \: number}{common \: denominator} }$

Difference of like fractions : $\mathsf{ \frac{difference \: of \: their \: numerator}{common \: denominator} }$

Addition and subtraction of unlike fractions

While performing the addition and subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fractions. Thus, obtained fractions should be reduced into lowest terms if there are any common on numerator and denominator.

For example:

$\mathsf{add \: \frac{1}{2} \: and \: \frac{1}{3} }$

L.C.M of 2 and 3 = 6

So, ⇒ $\mathsf{ \frac{1 \times 3}{2 \times 3} + \frac{1 \times 2}{3 \times 2} }$

⇒ $\mathsf{ \frac{3}{6} + \frac{2}{6} }$

⇒ $\frac{5}{6}$

Multiplication of fractions

To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term.

When any number or fraction is divided by a fraction, we multiply the dividend by reciprocal of the divisor. Let's consider a multiplication of a whole number by a fraction:

$\mathsf{4 \times \frac{3}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6}$

Multiplication for $\mathsf{ \frac{6}{5} \: and \: \frac{25}{3} }$ is done by the similar process

$\mathsf{ = \frac{6}{5} \times \frac{25}{3} = 2 \times 5 \times 10}$

Hope I helped!

Best regards!

5 0

2 years ago

Please help i don’t understand

choli [55]

Is that everything for the problem??

||-//

6 0

2 years ago

Class 607 has 27 students: 14 students have

LenaWriter [7]

Answer:

teal

Step-by-step explanation:

brown blue green and teal are the only eye colors out there

5 0

2 years ago

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

k0ka [10]

Let $a,b,c$ be the randomly selected lengths. Without loss of generality, suppose $a[tex]P(A + B \ge C) = P(A + B - C \ge 0)$

where $A,B,C$ are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

$P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)$

so that the joint density function is

$P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}$

where $[0,1]^3=[0,1]\times[0,1]\times[0,1]$ is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

$a + b - c = 0$

with $(a,b,c)\in\Bbb R^3$ . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region $a+b-c\ge0$ , since $0+0-1=-1$ . So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

$\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}$

5 0

1 year ago