All categories

3 years ago

The sum of two numbers is 8. When four times the smaller

Mathematics

2 answers:

Stolb23 [73]3 years ago

5 0

Answer:

x+x=8

that's mean: 2.x/4 + 4.x/1 = 18/1

we find at least a common multiple of 4 and multiply the units by 4

2x + 16x = 72

18x = 72/:18

x = 4

arsen [322]3 years ago

3 0

Answer: 7

suppose: the smaller number is a

the larger number is b

because The sum of two numbers is 8 => a + b = 8 (1)

four times the smaller number is added to two times the larger number, the result is 18 => 4a + 2b = 18 (2)

(1)(2) => a + b = 8

4a + 2b = 18

<=> a = 1

b = 7

=> the larger number is 7

Step-by-step explanation:

You might be interested in

What is 3 1/8 - 1 1/8 =

aalyn [17]

Answer:

Step-by-step explanation:

3 1/8 - 1/8 = 3

3-1 = 2

3 0

3 years ago

Read 2 more answers

A car travels along a straight stretch of road. It proceeds for 16.2 mi at 57 mi/h, then 24.6 mi at 44 mi/h, and finally 45.1 mi

LekaFEV [45]

Answer:

The average velocity the entire trip was 43.25 mi/h

Step-by-step explanation:

In this case we have to calculate an average based on the miles traveled. First we have to calculate the total of miles traveled, then calculate the portion of the total travel of each and with this calculate the average speed during the trip. First the total miles traveled:

$TotalMiles=16.2+24.6+45.1=85.9$

Now the percentages:

At 57 mi/h were $\frac{16.2}{85.9}=0.19$

At 44 mi/h were $\frac{24.6}{85.9}=0.29$

At 37.8 mi/h were $\frac{45.1}{85.9}=0.52$

Now multiplying the speed by the portion and summing them we can have the average velocity:

$57*0.19+44*0.29+37.8*0.52=43.25$

The average velocity the entire trip was 43.25 mi/h

5 0

3 years ago

Using fermat's little theorem, find the least positive residue of $2^{1000000}$ modulo 17.

torisob [31]

Fermat's little theorem states that
$a^p$ ≡a mod p

If we divide both sides by a, then
$a^{p-1}$ ≡1 mod p
=>
$a^{17-1}$ ≡1 mod 17
$a^{16}$ ≡1 mod 17

Rewrite
$a^{1000000}$ mod 17 as
$=(a^{16})^{62500}$ mod 17
and apply Fermat's little theorem
$=(1)^{62500}$ mod 17
=>
$=(1)$ mod 17

So we conclude that
$a^{1000000}$ ≡1 mod 17

6 0

4 years ago

PLEASE ANSWER + BRAINLIEST!! Factor completely. 4k - 20k^9 = 3b^2 - 108 =

Neporo4naja [7]

$4k-20k^9=4k(1-5k^8)=4k\left[1^2-(k^4\sqrt5)^2\right]\\\\=4k(1-k^4\sqrt5)(1+k^4\sqrt5)=4k\left[1^2-(k^2\sqrt[4]5)^2\right](1+k^4\sqrt5)\\\\=4k(1-k^2\sqrt[4]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k\left[1^2-(k\sqrt[8]5)^2\right](1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k(1-k\sqrt[8]5)(1+k\sqrt[8]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)$

$3b^2-108=3(b^2-36)=3(b^2-6^2)=3(b-6)(b+6)$

$Used:\ (a-b)(a+b)=a^2-b^2$

7 0

3 years ago

Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g

SpyIntel [72]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad B
\end{array}$

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.

8 0

3 years ago

Read 2 more answers