All categories

3 years ago

Please help with this question

Mathematics

2 answers:

Arlecino [84]3 years ago

7 0

Answer:

Option F

Step-by-step explanation:

<A is opposite to Side BC "only".

Semenov [28]3 years ago

5 0

Answer:

Line BC

Step-by-step explanation:

Opposite means whatever is exactly parallel to angle a.

In this case, we notice BC to satisfy that rule.

Hope this helps.

Good Luck

You might be interested in

The number a exceeds the number b by 50 percent. By what percent is the number b smaller than the number a?

inna [77]

Answer:

33 1/3 percent smaller

Step-by-step explanation:

The number a exceeds the number b by 50 percent.

a = b + b*.5

a = b(1.5)

Divide by 1.5

a/1.5 = b

2/3a =b

We want to find (a-b)/a to find the percent decrease

(a-2/3a)/a *100percent

(1/3a)/a* 100 percent

1/3 * 100 percent

33 1/3 percent smaller

5 0

2 years ago

compared to last year, the population of Boom Town has increased by 10%. The population is now 6,600. What was the population la

Rufina [12.5K]

Answer:

Step-by-step explanation:

6600 x 10% = 60&$;#;,$:#&#>$

6 0

2 years ago

Read 2 more answers

To make 2 batches of nut bars, Jayda needs to use 4 eggs.How many eggs are used in each batch of nut bars?

Charra [1.4K]

Answer:

Step-by-step explanation:

cause 2

7 0

3 years ago

Read 2 more answers

Simplify the expression 2x - 7 + 4x

bija089 [108]

Answer:

6x−7

Step-by-step explanation:

2x−7+4x

=2x+−7+4x

Combine Like Terms:

=2x+−7+4x

=(2x+4x)+(−7)

=6x+−7

3 0

3 years ago

Read 2 more answers

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago