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3 years ago

I NEED THIS DONE A.S.A.P. 50 POINTS!!!!!!

Mathematics

2 answers:

jarptica [38.1K]3 years ago

7 0

I was going to say the same thing but that right

MakcuM [25]3 years ago

5 0

Y+4=-3(X+1)
Point slope form can be written as
Y-k=M(X-p)
Where k is a Y value on the graph, and p is a x value on the graph and M is the slope. Note that p and k must be on the same point I.E (p,k)

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Use linear approximation to approximate √25.3 as follows.

Sophie [7]

The idea is to use the tangent line to $f(x)=\sqrt x$ at $x=25$ in order to approximate $f(25.3)=\sqrt{25.3}$ .

We have

$f(x)=\sqrt x\implies f(25)=\sqrt{25}=5$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}$

so the linear approximation to $f(x)$ is

$L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92$

Hence $m=\frac1{10}$ and $b=\frac92$ .

Then

$f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}$

4 0

3 years ago

Dan drank 7/8 of a bottle of water During basketball practice. He then drank Another 4/8 of a bottle after practice. How much wa

Lina20 [59]

Answer:

In total Dan dranked 11/8 of his bottle or as a mixed number 1 and 3/8 of his bottle.

Step-by-step explanation:

6 0

3 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

3 years ago

Which expression represents h(x)?

blsea [12.9K]

Answer:

(g-f)x is the answers for the question

Step-by-step explanation:

please give me brainlest and follow me

3 0

3 years ago

Solve: (a)(9a + 4) = 0.

Monica [59]

Answer:

a = -4/9

Step-by-step explanation:

(9a + 4) = 0

Subtract 4 from each side

9a + 4-4 = 0-4

9a = -4

Divide each side by 9

9a/9 = -4/9

a = -4/9

4 0

3 years ago

Read 2 more answers