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2 years ago

13% of $99.99 = $26.00 O $25.98 O $13.00 O $12.99

Mathematics

1 answer:

vova2212 [387]2 years ago

8 0

Answer:

12.99 or 13 rounded

Step-by-step explanation:

99.99/(.13)=12.9987

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10x+36-38x-47 I don't understand this question i am so confused please explain i would be really thankful.

Lelechka [254]

-28x-11

Step-by-step explanation:

10x-38x=-28x

+36-47=-11

so the answer is -28x-11

8 0

3 years ago

On Earth day, a local community wants to distribute 800 flyers, 300 banners, and 250 badges to volunteers in packets containing

steposvetlana [31]

Answer:

800 flyersOn Earth day, a local community wants to distribute 800 flyers, 300 banners, and 250 badges to volunteers in packets containing the three items. What is the greatest number of packets that can be made using all these items if each type of item is equally distributed among the packets?

4 0

2 years ago

Explain how to answer the question attached

LenaWriter [7]

Answer:

The players ran 150 meters

Step-by-step explanation:

Pythagorean Theorem: a² + b² = c²

Since we have a rectangle with sides 90 and 120, we know if we split a diagonal across it, we will get a right triangle with legs 90 and 120. From there, we use Pythagorean Theorem to solve:

90² + 120² = c²

8100 + 14400 = c²

c² = 22500

√c² = √22500

c = 150

3 0

3 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

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 = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

Classify triangle ABC by its sides if side AB = 6x, side AC = 4x + 6 and side BC = 8x + 3. The perimeter of triangle ABC is 63.

sladkih [1.3K]

6x + 4x+6 + 8x+3 = 63
6x+4x+8x = 63 - 3 - 6
18x = 54
x = 54/18
x = 3

Side AB = 6(3) = 18
Side AC = 4(3)+6 = 18
Side BC = 8(3)+3 = 27

3 0

3 years ago

Read 2 more answers