All categories

3 years ago

What property is 7+(4+6)=7+n

Mathematics

1 answer:

Ket [755]3 years ago

7 0

This looks like "half" a property. I'm assuming you mean something that looks like this:

IF a + b = a + c, THEN b = c. In the expression in your problem, the role of a is being played by 7. The role of b is played by (4 + 6), and the role of c is played by n.

The conclusion would be that 4 + 6 = n.

But you asked about what the property is called. The likely name is the subtraction property of equality. How did we go from a + b = a + c to the conclusion b = c? We subtracted a from both sides of the equation.

You might be interested in

PLEASE HELP ME I WILL HIVE BRAINLIST TO THE CORRECT ANSWER !!!!!!!!!!!

astra-53 [7]

Answer:

I'm not 100 percent sure but I think it's the first answer choice because at 40000 3 years have passed. Hope this helps :)

5 0

3 years ago

Read 2 more answers

The angles of a triangle are 42º, 63º, xº.<br><br> What is the value of x?

MakcuM [25]

75 degrees because 42+63=105
180-105=x

6 0

4 years ago

Read 2 more answers

I need 10x+3y=19 and y=2x+5 solved algebraically

taurus [48]

Y=2x+5
sub 2x+5 for y in other equaiton
10x+3(2x+5)=19
distribute
10x+6x+15=19
minus 15 from both sides
16x=4
divide both sides by 16
x=1/4
sub back

y=2x+5
y=2(1/4)+5
y=1/2+5
y=5.5=11/2

solution is (1/4,11/2) or in decimal, (0.25,5.5)

8 0

4 years ago

During the 2015-16 NBA season, J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 . Assume th

ser-zykov [4K]

Answer: 0.5898

Step-by-step explanation:

Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .

We assume that,

The probability that .J. Redick makes any given free throw =0.901 (1)

Free throws are independent.

So it is a binomial distribution .

Using binomial probability formula, the probability of getting success in x trials :

$P(X=x)^nC_xp^x(1-p)^{n-x}$

, where n= total trials

p= probability of getting in each trial.

Let x be binomial variable that represents the number of a=makes.

n= 14

p= 0.901 (from (1))

The probability that he makes at least 13 of them will be :-

$P(x\geq13)=P(x=13)+P(x=14)$

$=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898$

∴ The required probability = 0.5898

5 0

3 years ago

Simplify the square root of four over the cubed root of four.

Vikki [24]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\\\ ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\sqrt{4}}{\sqrt[3]{4}}\implies \cfrac{\sqrt[2]{4}}{\sqrt[3]{4}}\implies \cfrac{4^{\frac{1}{2}}}{4^{\frac{1}{3}}}\implies 4^{\frac{1}{2}}\cdot 4^{-\frac{1}{3}}\implies 4^{\frac{1}{2}-\frac{1}{3}}\implies 4^{\frac{3-2}{6}} \\\\\\ 4^{\frac{1}{6}}\implies (2^2)^{\frac{1}{6}}\implies 2^{2\cdot \frac{1}{6}}\implies 2\frac{1}{3}\implies \sqrt[3]{2}$

7 0

3 years ago

Read 2 more answers