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2 years ago

Which equation supports the given conjecture?

2 answers:

7 0

The verified answer is wrong the correct answer is d 5 squared plus 6 squared. Because the question is asking for two consecutive integers and it wants an odd integer therefor d is the correct answer because it has to squared consecutive integers and an odd answer.

Hope i helped Mark as brainliest

Readme [11.4K]2 years ago

3 0

The first part says the sum of the squares of two consecutive integers or in other words (x)(x)+(x+1)(x+1) so we can cross out A because 2 and 3 are prime numbers (no factors) we can cross out all the others because the square root of 52 is not an integer so any equation with 52 in it does not satisfy the requirement. so none of them are corect. additionally, some of the equations are obviously false such as 52+62=61

You might be interested in

How to find variables of these #2 & #3

Lubov Fominskaja [6]

So... hmmm if you check the first picture below, for 2)

we could use the proportions of those small, medium and large similar triangles like $\bf \cfrac{small}{large}\qquad \cfrac{x}{12}=\cfrac{6}{x}\impliedby \textit{solve for "x"}
\\\\\\
\cfrac{small}{large}\qquad \cfrac{z}{18}=\cfrac{6}{z}\impliedby \textit{solve for "z"}
\\\\\\
\cfrac{large}{medium}\qquad \cfrac{y}{12}=\cfrac{18}{y}\impliedby \textit{solve for "y"}$

now.. for 3) will be the second picture below

$\bf \cfrac{large}{medium}\qquad \cfrac{x+10}{2\sqrt{30}}=\cfrac{2\sqrt{30}}{10}\impliedby \textit{solve for "x"}
\\\\\\
\textit{now, because you already know what "x" is, we can use it below}
\\\\\\
\cfrac{large}{small}\qquad \cfrac{z}{x}=\cfrac{x+10}{z}\impliedby \textit{solve for "z"}
\\\\\\
\textit{and let us use "x" again below}
\\\\\\
\cfrac{small}{medium}\qquad \cfrac{y}{10}=\cfrac{x}{y}\impliedby \textit{solve for "y"}$

8 0

2 years ago

Plz help its so confusing

aleksklad [387]

Answer:

20, 4

Step-by-step explanation:

Let's assume total number of students registered is n

Last week n/4 people are absent

so people present are n - n/4 = <u>3n/4</u>

Today after return of Jen 1/5 n are absent

so people present are n - n/5 = 4n/5

which is also equal to last week present people + 1 (Jen)

= 3n/4 + 1

= (3n + 4) / 4

so (3n + 4) / 4 = 4n/5

=> 3n + 4 = 4n/5 * 4

=> (3n + 4) * 5 = 16n

=> 15n + 20 = 16n

=> 20 = n

=> n = 20

So total students registered for class is 20

today no of people absent = n/5 = 20/5 = 4

8 0

2 years ago

Answer please! I have been "getting" the correct answer, as far as I know. The answer which I got was 10x^2 - 4x. But I don't se

GarryVolchara [31]

Answer:

The answer is 10x² - 4x.

Step-by-step explanation:

You have to apply Distributive Law,

$a(m + n) = am + an$

So for this question :

$2x(5x - 2)$

$= 2x(5x) + 2x( - 2)$

$= 10 {x}^{2} - 4x$

4 0

3 years ago

Give an example of a function f : N → N that is surjective but not injective. You must explain why your example is surjective an

Helen [10]

Answer:

Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.

Step-by-step explanation:

First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y. We conclude that f is surjective.

However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.

Note:

If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.

7 0

2 years ago

Un apostador pierde en su primer juego el 30% de su dinero, en el segundo juego pierde el 50% de lo que perdió; finalmente en el

sergiy2304 [10]

Responder:

26,62

Explicación paso a paso:

Sea x el dinero original que tenía el jugador:

si un jugador pierde en su primer juego el 30% de su dinero, la cantidad perdida será;

$= 30/100 \ of \ x\\= 0.3x$

Si en el segundo juego pierde el 50% de lo que perdió, entonces la cantidad perdida en el segundo juego será:

$= 50/100 \ of \ 0.3x\\= 0.5 \times 0.3x\\= 0.15x$

Si en el tercer juego pierde el 40% de todo lo que ha perdido, la cantidad perdida en el tercer juego será:

$=\frac{40}{100} \ of \ (0.3x+0.15x) \\= 0.4(0.45x)\\= 0.2025x$

Si la cantidad que le queda para seguir apostando es de 37 soles, entonces para calcular la cantidad original que tiene, sumaremos toda la cantidad perdida y la cantidad restante y equipararemos la cantidad original x como se muestra:

0,3x + 0,15x + 0,2025x + 37 = x

0,6525x + 37 = x

x-0,6525x = 37

0,3475x = 37

x = 37 / 0,3475

x = 106,48

La cantidad que tenía originalmente era de 106,48

75% de 106,48

= 75/100 * 106,48

= 0,75 * 106,48

= 79,86

Tomando la diferencia entre su monto original y su 75% será:

$= 106.48-79.86\\= 26.62$

3 0

2 years ago