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3 years ago

8

Determine the number of significant figures in each of the following numbers

1 answer:

Strike441 [17]3 years ago

6 0

Answer:

9.99999999999999 apk

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Please help me on this problem

nevsk [136]

a cause their really isnt any ADEF like in letters

7 0

3 years ago

Graph the system & write its solution <br> 2x+y=-4<br> Y=-1/2x-1

xz_007 [3.2K]

Answer:

The solution of system of equation is (-2,0)

Step-by-step explanation:

Given system of equation are

Equation 1 : 2x+y=(-4)

Equation 2 : y+ $\frac{1}{2}$ x=(-1)

To plot the equation of line, we need at least two points

For Equation 1 : 2x+y=(-4)

Let x=0

2x+y=(-4)

2(0)+y=(-4)

y=(-4)

Let x=1

2x+y=(-4)

2(1)+y=(-4)

y=(-6)

Therefore,

The required points for equation is (0,-4) and (1,-6)

For Equation 2 : y+ $\frac{1}{2}$ x=(-1)

Let x=0

y+ $\frac{1}{2}$ x=(-1)

y+ $\frac{1}{2}$ (0)=(-1)

y=(-1)

Let x=2

y+ $\frac{1}{2}$ x=(-1)

y+ $\frac{1}{2}$ (2)=(-1)

y=(-2)

The required points for equation is (0,-1) and (2,-2)

Now, plot the graph using this points

From the graph,

The red line is equation 1 and blue line is equation 2

Since. The point of intersection is solution of system of equations

The solution of system of equation is (-2,0)

6 0

3 years ago

Please this a really easy question. Does this mean I got an A on this test?! Please answer!!!!!!!!!!!!!!!

AnnZ [28]

Answer:

yes it does. good job on your A.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Can someone please help me??

dmitriy555 [2]

Answer:

I is clear that, the linear equation $5x+12=5x-7$ has no solution.

Step-by-step explanation:

<u>Checking the first option:</u>

$\frac{2}{3}\left(9x+6\right)=6x+4$

$6x+4=6x+4$

$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$

$6x+4-4=6x+4-4$

$6x=6x$

$\mathrm{Subtract\:}6x\mathrm{\:from\:both\:sides}$

$6x-6x=6x-6x$

$0=0$

$\mathrm{Both\:sides\:are\:equal}$

$\mathrm{True\:for\:all}\:x$

<u>Checking the 2nd option:</u>

$5x+12=5x-7$

$\mathrm{Subtract\:}5x\mathrm{\:from\:both\:sides}$

$5x+12-5x=5x-7-5x$

$\mathrm{Simplify}$

$12=-7$

$\mathrm{The\:sides\:are\:not\:equal}$

$\mathrm{No\:Solution}$

<u>Checking the 3rd option:</u>

$4x+7=3x+7$

$\mathrm{Subtract\:}7\mathrm{\:from\:both\:sides}$

$4x+7-7=3x+7-7$

$\mathrm{Simplify}$

$4x=3x$

$\mathrm{Subtract\:}3x\mathrm{\:from\:both\:sides}$

$4x-3x=3x-3x$

$\mathrm{Simplify}$

$x=0$

<u>Checking the 4th option:</u>

$-3\left(2x-5\right)=15-6x$

$\mathrm{Subtract\:}15\mathrm{\:from\:both\:sides}$

$-6x+15-15=15-6x-15$

$\mathrm{Simplify}\$

$-6x=-6x$

$\mathrm{Add\:}6x\mathrm{\:to\:both\:sides}$

$-6x+6x=-6x+6x$

$\mathrm{Simplify}$

$\mathrm{Both\:sides\:are\:equal}$

$\mathrm{True\:for\:all}\:x$

Result:

Therefore, from the above calculations it is clear that, the linear equation

$5x+12=5x-7$ has no solution.

4 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago