All categories

3 years ago

Write 26 as repeated multiplication?

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

5 0

Answer:

This is your answer

Table of 26 gives the repeated addition of number 26, repeated for a certain number of times. For example, 4 baskets of 26 apples each, gives the sum of 26 + 26 + 26 + 26 = 104.

You might be interested in

Write the ordered pair that represents YZ. Then find the magnitude of YZ. y(-4,12), Z(1,19).

ki77a [65]

The magnitude of YZ is 8.6

Explanation:

Y( -4, 12)

Z ( 1, 19)

Magnitude of YZ = ?

We know:

$YZ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

On substituting the value we get:

$YZ = \sqrt{(1+4)^2 + (19 - 12)^2} \\\\YZ = \sqrt{(5)^2 + (7)^2} \\\\YZ = \sqrt{25+49} \\\\YZ = \sqrt{74} \\\\YZ = 8.6$

Thus, the magnitude of YZ is 8.6

6 0

3 years ago

Read 2 more answers

Combine like terms. 2y2 – 5+1 + 2y2 – 6y2 – 3 – 4y ASAP someone help please

AnnZ [28]

Answer:

-2y2 - 4y - 7

Step-by-step explanation:

2y2 + 2y2 - 6y2 - 4y - 5 - 3 + 1

4y2 - 6y2 - 4y - 8 + 1

-2y2 - 4y - 7

3 0

3 years ago

Read 2 more answers

The volume of a sphere is 500 3 πcm3. What is the radius? Sphere V = 4 3 πr3 1. Substitute value into formula: 500 3 π = 4 3 πr

Kisachek [45]

Answer:

Radius of the sphere is 5m.

Step-by-step explanation:

We have given the volume of sphere $\frac{500}{3}\cdot {\pi}m^3$

we have to find the radius of the sphere

$\text{volume of sphere is }=\frac{4}{3}\cdot {\pi}\cdot r^3$

On substituting the given value of volume of sphere we get

$\frac{500}{3}\cdot {\pi}=\frac{4}{3}\cdot {\pi}\cdot r^3$

After simplification we get:

$\Rightarrow 4\cdot r^3=500$

$\Rightarrow r^3=125$

$\Rightarrow r=5$

Hence, the radius of sphere is 5m

7 0

3 years ago

Read 2 more answers

ABCD ~ QRST

slava [35]

The answer for this question is 4.

5 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

Write 26 as repeated multiplication?​

Write 26 as repeated multiplication?