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oksano4ka [1.4K]

3 years ago

11

Circle the vertex of the angle.

1 answer:

Zina [86]3 years ago

5 0

Answer: B

Step-by-step explanation:

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A dealer selects a card at random from a standard deck of 52 cards.

Oksanka [162]

Answer:

a) you are correct about a.

b)2/13 chance

c) 1/2 chance

please mark me as brainliest and give me a thanks and five stars please

3 0

3 years ago

Which equation is equivalent to √x^2+81 = x+10

jeka57 [31]

Keywords:

<em>equation, operations, equivalent, binomial, square root </em>

For this case we have an equation in which we must apply operations to rewrite it in an equivalent way. We must start by raising both sides of the equation to the square. Thus, we eliminate the square root of the left side of equality and finally solve the binomial of the right side of equality.

So we have:

$\sqrt {x ^ 2 + 81} = x + 10\\(\sqrt {x ^ 2 + 81}) ^ 2 = (x + 10) ^ 2\\x ^ 2 + 81 = (x + 10) ^ 2$

By definition: $(a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2$

$x ^ 2 + 81 = x ^ 2 + 2 (x) (10) + (10) ^ 2\\x ^ 2 + 81 = x ^ 2 + 20x +100$

Thus, $x ^ 2 + 81 = x ^ 2 + 20x +100$ is equivalent to $\sqrt {x ^ 2 + 81} = x + 10$

Answer:

$x ^ 2 + 81 = x ^ 2 + 20x +100$

Option D

4 0

4 years ago

Read 2 more answers

Need this answered please!!

Paraphin [41]

<h2>ABC and DEF</h2>

Their 3 angles are equal (64,26,90)

Hope this will help...

Please mark as brainliest

6 0

3 years ago

What is the answer for the problem?

TEA [102]

Answer:

A: I assume that the equation 4x+2 has a greater rate of change.

B: The function F has a greater y intercept.

Step-by-step explanation:

Trend in the table is multiply the x by 3 then add 6.

Your equation would be y = 3x+6 for the table.

6 0

3 years ago

Find two positive consecutive odd integers such that square of the smaller integer is 10 more than the larger integer

Otrada [13]

Let $2n+1$ be the smaller integer. The larger integer is then $2n+3$ , and we have

$(2n+1)^2=10+(2n+3)\implies4n^2+4n+1=2n+13$

$\implies4n^2+2n-12=0$

$\implies2n^2+n-6=0$

$\implies(2n-3)(n+2)=0$

$\implies 2n-3=0\text{ or }n+2=0$

$\implies n=\dfrac32\text{ or }n=-2$

We omit $n=-2$ , since $2(-2)+1=-3$ is negative.

Then for $n=\dfrac32$ we find $2\left(\dfrac32\right)+1=4$ , but this is not odd.

There are no consecutive odd integers that satisfy the given condition!

3 0

3 years ago