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3 years ago

Solve the equation 4=t/2.5 t=

Mathematics

2 answers:

julia-pushkina [17]3 years ago

8 0

Answer:

Step-by-step explanation:

Multiply 4 by 2.5 to get 10

ELEN [110]3 years ago

6 0

Answer:

Step-by-step explanation:

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Which of the following expressions is equivalent to 13^(2)-3^(2)

Amiraneli [1.4K]

Answer:

160

Step-by-step explanation:

13² - 3² = 13 × 13 - 3 × 3 = 169 - 9 = 160

8 0

2 years ago

I need help please!!!!!!!

k0ka [10]

Answer:

1 roll of ribbon, 1 package of buttons, and 3 packages of beads

Step-by-step explanation:

Use the information you found about how much each makes to answer the question.

Lynette will make 6 decorations.

Since 1 roll makes 6, she needs 1 roll.

Since 1 package of buttons makes 6, she needs 1 package.

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5 0

3 years ago

Tony collected 16.2 pounds of pecans from the trees on his farm. He will give the same weight of pecans to each of 12 friends. H

netineya [11]

Answer:

1.35 pounds each

Step-by-step explanation:

16.2 divided by 12 equals 1.35

6 0

3 years ago

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

A shop owner wants to market a new vest. The vest distributor will charge a reduced price for one size of the owner’s choice. Th

AlladinOne [14]

The answer is 44 because the definition of Mode is the number that appears the most. The correct answer is A

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4 years ago