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3 years ago

I need an answer and explanation to this please help.

Mathematics

1 answer:

Radda [10]3 years ago

3 0

$\displaystyle\\
64\times \frac{3}{8} = \frac{64\times 3}{8} =64:8\times3= 8 \times 3 = \boxed{\bf24 }$

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Find the sum of the first 30 multiples of 8.

Anna11 [10]

Step-by-step explanation:

sum of the arithmetic sequence

a1 = 8

a2 = a1 + 8 = 8+8 = 16

...

an = 8×n

the sum of the first n elements of such a series is

n/2 × (2×a1 + (n - 1)×d]

with d being the difference between 2 consecutive elements (= the added constant).

so,

s30 = 30/2 × (2×8 + 29×8) = 15×(31×8) = 3720

3 0

2 years ago

Please helppp number 11

aivan3 [116]

Answer:

(1/3)⁷

Step-by-step explanation:

negative exponents work like this:

x⁻²=(1/x)²

Apply this theory to this equation:

3⁻⁷=

(1/3)⁷

4 0

3 years ago

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Naddik [55]

Answer:

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Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The slope of the line below is -3. Which of the following is the point-slope

Anna [14]

Given:

The slope of the line is -3.

The line passes through the point (2,-2).

To find:

The point-slope form of the line.

Solution:

Point slope form: If a line passes through the point $(x_1,y_1)$ with slope m, then the point-slope form of the line is:

$y-y_1=m(x-x_1)$

The slope of the line is -3 and it passes through the point (2,-2). So, the point-slope form of the line is:

$y-(-2)=-3(x-2)$

$y+2=-3(x-2)$

Therefore, the required point slope form of the given line is $y+2=-3(x-2)$ .

5 0

3 years ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

<u>Step-by-step explanation:</u>

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

3 0

3 years ago