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3 years ago

I WILL GIVE BRAINLEST PLEASE HELP

Mathematics

1 answer:

guapka [62]3 years ago

7 0

Answer:

Step-by-step explanation:

In this question, we are concerned with selecting which of the options best represents the difference of two squares.

Let’s have an exposition below as follows;

Consider two numbers, which are perfect squares and can be expressed as a square of their square roots;

a^2 and b^2

where a and b represents the square roots of the numbers respectively.

Inserting a difference between the two, we have;

a^2 - b^2

Now by applying the difference of two squares, these numbers will become;

a^2 - b^2 = (a + b)(a-b)

So our answer out of the options will be that option that could be expressed as above.

The correct answer to this is option A

Kindly note that;

x^2 -9 can be expressed as x^2 - 3^2 and consequently, this can be written as;

(x-3)(x + 3)

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8. In AABC, LA is a right angle and mzB = 45°. Find BC.

Margaret [11]

Answer:

BC = 11 $\sqrt{2}$

Step-by-step explanation:

using the sine ratio in the right triangle and the exact value

sin45° = $\frac{1}{\sqrt{2} }$ , then

sin 45° = $\frac{opposite}{hypotenuse}$ = $\frac{AC}{BC}$ = $\frac{11}{BC}$ = $\frac{1}{\sqrt{2} }$ ( cross- multiply )

BC = 11 $\sqrt{2}$

7 0

2 years ago

The pattern 7, 21, 63, 189, ________ follows the rule multiply by 3. What is the next number in the pattern?

Radda [10]

Answer

The answer is 567

Step-by-step explanation:

189*3=567

6 0

2 years ago

Lines g, h, and l are parallel and m< 2 = 132°.

xxMikexx [17]

M < 2 and m < 6 are corresponding angles and are equal....so < 6 = 132.

< 6 and < 8 form a line and are equal to 180

< 6 + < 8 = 180
132 + < 8 = 180
< 8 = 180 - 132
< 8 = 48 <===

4 0

3 years ago

Read 2 more answers

Someone please help me ASAP

Minchanka [31]

Answer:

There is 1 boy per 2 girls

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

5 0

3 years ago