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2 years ago

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Mathematics

1 answer:

goldfiish [28.3K]2 years ago

3 0

Answer: the answer is 80. Hope this helps

Step-by-step explanation:

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6×10(6)+7×10(5)+3×10(2)+8×10+2

anastassius [24]

Answer:

852

6×10(6)+7×10(5)+3×10(2)+8×10+2= 852

Step-by-step explanation:

ORDER OF OPERATIONS Will really help you solve this long problem.

Order of operations is PEMDAS.

P-parenthesis

E-exponents

M-multiplication

D-division

A-addition

S-subtraction

6)(10)(6)+(7)(10)(5)+(3)(10)(2)+(8)(10)+2

=(60)(6)+(7)(10)(5)+(3)(10)(2)+(8)(10)+2

=360+(7)(10)(5)+(3)(10)(2)+(8)(10)+2

=360+(70)(5)+(3)(10)(2)+(8)(10)+2

=360+350+(3)(10)(2)+(8)(10)+2

=710+(3)(10)(2)+(8)(10)+2

=710+(30)(2)+(8)(10)+2

=710+60+(8)(10)+2

=770+(8)(10)+2

=770+80+2

=850+2

=852

7 0

2 years ago

Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th

zalisa [80]

We have the following equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8$

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

$y \leq -0.5x+5 \\ y \leq -1.25x+8$

Region B.

Let's take a point that is in this region, that is:

$P(0,6)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \leq -1.25x+8$

Region C.

A point in this region is:

$P(0,10)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \ \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Region D.

A point in this region is:

$P(8,0)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \ \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2$

So the inequalities are:

$(1) \ y \leq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\ \\ Solving \ for \ y: \\ y=-0.5(4)+5=3$

Accordingly, the point is:

$\boxed{p(4,3)}$

Point q:

This point is the $x-intercept$ of the line:

$y=-0.5x+5$

So let:

$y=0$

Then

$x=\frac{5}{0.5}=10$

Therefore, the point is:

$\boxed{q(10,0)}$

Part b)

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

$O(0,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}$

It is true.

Region B.

Let's take a point that is in this region, that is:

$P(0,7)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}$

It is true

Region C.

Let's take a point that is in this region, that is:

$P(0,11)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}$

It is true

Region D.

Let's take a point that is in this region, that is:

$P(9,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}$

It is true

7 0

3 years ago

Prove: lim x^3 = 8. x approaches 2

Maslowich

When we compute the value of a limit, all that we do is to change the value of the parameter
lim x^3 = (2)^3= 2x2x2=8
x approaches 2

5 0

2 years ago

A particle is moving along the parabola x^2=4(y+6). As the particle passes through the point (8,10), the rate of change of its y

Fofino [41]

You'll need to use differentiation (specifically, implicit differentiation) here.

If x^2 = 4(y+6), differentiating both sides with respect to time t produces the following:

2x (dx/dt) = 4([dy/dt]) (note that (d/dt) 6 = 0)

We need to solve for (dx/dt). Substitute 8 for x (y does not appear in this latest equation, so we do nothing with y=10). Substitute the given 5 units/sec for dy/dt:

2(8)(dx/dt) = 4(5)(units/sec)

Solving for dx/dt, dx/dt = [20 units/sec]/16, or 5/4 units/sec, or 1.25 units/sec.

8 0

3 years ago

What is the slope of the line -9, -6 and 3, -9

adell [148]

Answer:

Step-by-step explanation:

We find the slope is 1/4 by graphing the points. Then counting the slope as rise/run

5 0

2 years ago