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2 years ago

7

Two four digit numbers abcd and efgh are constructed using the digits from 1 to 8. [a. b. c. d. e. f. g and h are the digits of

the two four digit numbers] It is found that abcd x 2 = efgh. Find the numbers.

1 answer:

NeX [460]2 years ago

4 0

Answer:

that is your answer on the picture

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Mr North spent $144,00 to build a fence around the perimeter of bis vegetable garden.He paid $6.oo per yard for fencing a, Draw

DedPeter [7]

The perimeter of his fence is 2400 the steps i took to find this was i divided 14400 by 6 an i got 2400. I can only think of one possible reason I hope the helps, Sorry

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3 years ago

I like to give people easy points so here: 2 + 4

Brilliant_brown [7]

Answer:

6

Step-by-step explanation:

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3 years ago

Read 2 more answers

Solve for the indicated variable.<br><br> 2a + 2b = c for b<br><br> B = ?/?

kodGreya [7K]

Answer:

b = c − 2a

------------

2

Step-by-step explanation:

2a + 2b = c

2b = c − 2a

b = c − 2a

------------

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7 0

3 years ago

56.27 in expanded form

Lana71 [14]

Answer:

Step-by-step explanation:

56.27 =

50

+ 6

+ 0.2

+ 0.07

Expanded Factors Form:

56.27 =

5 × 10

+ 6 × 1

+ 2 × 0.1

+ 7 × 0.01

Expanded Exponential Form:

56.27 =

5 × 101

+ 6 × 100

+ 2 × 10-1

+ 7 × 10-2

Word Form:

56.27 =

fifty-six and twenty-seven hundredths

5 0

2 years ago

Rewrite in Polar Form....<br> x^(2)+y^(2)-6y-8=0

skad [1K]

$\bf x^2+y^2-6y-8=0\qquad 
\begin{cases}
x^2+y^2=r^2\\\\
y=rsin(\theta)
\end{cases}\implies r^2-6rsin(\theta)=8
\\\\\\
\textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8
\\\\\\\
[r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby 
\begin{array}{llll}
\textit{so we need a value there to make}\\
\textit{a perfect square trinomial}
\end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there}
\\\\\\
\textit{now, bear in mind we're just borrowing from zero, 0}
\\\\
\textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\
\left. \qquad \right.\uparrow \\
\textit{so-called "completing the square"}
\\\\\\\
[r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta)
\\\\\\
r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$

3 0

3 years ago