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3 years ago

Which of the following is a true statement?

Mathematics

2 answers:

Zolol [24]3 years ago

8 0

Yea, the only answer would be D because it is a strait-forward answer the rest are unsure or half way.

nadya68 [22]3 years ago

5 0

It's D. That is the only true statement.

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∠A and \angle B∠B are vertical angles. If m\angle A=(6x-11)^{\circ}∠A=(6x−11) ∘ and m\angle B=(5x+4)^{\circ}∠B=(5x+4) ∘ , then f

ICE Princess25 [194]

Answer: $\angle B=79^{\circ}.$

Step-by-step explanation:

When two lines cross each other , then the opposite angles are known as vertical angles.

Vertical angles are equal.

Here, ∠A and ∠B are vertical angles.

So, ∠A = ∠B

If $\angle A=(6x-11)^{\circ}$ and $\angle B=(5x+4)^{\circ}$

Then , $6x-11=5x+4$

Subtract 5x from both sides, we get

$x-11=4$

Add 11 on both sides, we get

$x=15$

Now , $\angle B=(5(15)+4)^{\circ}=79^{\circ}$

Hence, the measure of $\angle B=79^{\circ}.$

3 0

2 years ago

I have attached the question for you

scoundrel [369]

Answer:

s = $\frac{1}{3}$

Step-by-step explanation:

s = ut + $\frac{1}{2}$ at² ( substitute the given values into the equation )

s = ( 3 × $\frac{1}{3}$ ) + ( $\frac{1}{2}$ × - 12 × ( $\frac{1}{3}$ )² )

= 1 + (- 6 × $\frac{1}{9}$ )

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

= 1 - $\frac{2}{3}$

= $\frac{1}{3}$

4 0

2 years ago

Read 2 more answers

For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;

Dmitrij [34]

Answer:

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

$ax+by=c$

Suppose integers x0 and yo satisfy the equation; that is,

$ax_0+by_0 = c$

what other values

$x = x_0+h$ and $y=y_0+k$

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

$as+bt=gcd(a,b)$

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

$a(sk) + b(tk) = gcd(a,b)k = c$

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

$a(x_1-x) + b(y_1-y)=0$

Therefore,

$a(x_1-x) = b(y-y_1)$

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

$y = y_1+r(\frac{a}{gcd(a, b)})$ for some integer r. Substituting into the equation

$a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba$

$x = x_1-r(\frac{b}{gcd(a, b)} )$

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

$15 = 6*2+3\\6=3*2+0$