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3 years ago

How do you graph y=2x-10

1 answer:

6 0

One of the easier approaches would be that of x- and y-intercepts.

Setting x=0 results in y=-10; the y-intercept is (0, -10).
Setting y = 0 results in x = 5, so the x-intercept is (5,0).

Plot these 2 points. Then draw a str. line thru these points.

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What is 83.34 written in expanded form?

Vilka [71]

Answer:

80+3+0.3+0.04

Step-by-step explanation:

8 0

2 years ago

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About how many cubes were used to make this figure? A. about 40 B. about 70 C. about 100 D. about 140

natta225 [31]

Answer:

Around 100,i.e. 105

Step-by-step explanation:

In the given cube

Length of the cube = 7 unit cells

Breadth of the cube = 3 unit cells

Height of the cube = 5 unit cells

Therefore the number of unit cubes required to make such big cube is nothing but the volume of the big cube = length*breadth*height

⇒Number of cubes used to make that big cube= 7*5*3

= 105

Hence, option D (around 100) is the correct answer

4 0

2 years ago

You panit 1/3 of a wall in 1/4 hour at that rate how long will it take you to paint one wall

kodGreya [7K]

Answer:

3/4 hour or 45 minutes

It will take 0.75 hours or 45 minutes or 2700 seconds to paint one wall according to the rate we're given.

7 0

2 years ago

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Pls give me an answer to this!!!!!!!

gayaneshka [121]

Answer:

CONSTRUCT FROM A TO B AND ALSO A TO C AND B TO C

5 0

2 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

2 years ago