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2 years ago

Can someone help me answer these two questions

Mathematics

1 answer:

Olin [163]2 years ago

8 0

Answer:

Step-by-step explanation:

for the 1st one you multiply 2+2 by 2 the answer is 8 i cant see the whole 2nd problem so I dont know

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3p-(-20)=50 what is p?

zlopas [31]

3p-(-20)=50

3p+20=50 subtract 20 from both sides

3p=30 divide both sides by 3

p=10

3 0

3 years ago

Read 2 more answers

In the metric system, multiplying by 1000 would be the same as moving the decimal point _____ place(s) to the right.

hodyreva [135]

It would be the same as moving the decimal three to the right, and that applies to everything, not just the metric system.

3 0

2 years ago

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Solve the following linear equation for k.<br> 2 (k-5) + 3k = k + 6

kiruha [24]

Answer:

2(k-5)+3k=k+6

-3k. -3k

2(k-5)=-2k+6

/2. /2

k-5=-k+3

+k. +k

2k-5=3

+5. +5

2k=8

/2. /2

k=4

7 0

2 years ago

A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that co

White raven [17]

Answer:

The value of x would be $\frac{3}{5}y$ or $\frac{3}{8}z$

Step-by-step explanation:

Given,

x tons of 10% gravel G mixture is mixed with y ton of 2% gravel G to obtain 5% gravel mixture,

Thus, gravel G in x ton + gravel in y ton = gravel G in the resultant mixture,

⇒ 0.1x + 0.02y = 0.05(x+y),

0.1x - 0.05x = 0.05y - 0.02y

0.05x = 0.03y

$\implies x = \frac{0.03}{0.05}y=\frac{3}{5}y----(1)$

Now, resultant mixture is z ton,

i.e. $z=x+y$

$\implies y = z - x$

From equation (1),

$x=\frac{3}{5}(z-x) = \frac{3}{5}z-\frac{3}{5}x$

$x+\frac{3}{5}x=\frac{3}{5}z$

$\frac{8}{5}x=\frac{3}{5}z$

$x=\frac{3}{8}z$

3 0

3 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

2 years ago