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2 years ago

Im in need of a big brain

Mathematics

1 answer:

stiv31 [10]2 years ago

8 0

Answer:

x = 34

Step-by-step explanation:

If it is bisected, that means the angles are equal <ABD = < CBD

The triangles are equal by SAS

That means AD = CD

2x+30 = 3x-4

subtract 2x from each side

2x+30 -2x = 3x-4 -2x

30 = x-4

Add 4 to each side

3x+4 = x-4+4

34 =x

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Georgia ,a florist,charges $10 per flower bundle plus a$15 delivery charge per order what is the slopes

andrew11 [14]

Answer:

The slope is $ 10 per flower bundle.

Explanation:

The slope of a linear function is the ratio of the change in the dependent variable over the change in the independent variable.

In the problem, the independent variable is the number of flower bundles and the dependent variable is the amount charged.

If you call n the number of flower bundles and c(n) the amount charged per order, the amount charged per order is modeled by linear function:

c(n) = 10n + 15

Where, 10 is the amount charged in dolars per flower bundle, and 15 is the delivery charge per order.

Then, for every increment in the number of flower bundles (n) the amount charged will increase in $ 10. That is the slope

slope = $ 10 / flower bundle.

3 0

3 years ago

The approximation of which irrational number is shown by point R on the number line?

Alenkasestr [34]

Answer:

B it is very simple my friend

Step-by-step explanation:

3.75 × 3.75 = 14.0

6 0

3 years ago

I need help ASAP! It's urgent.. PLISSSSS

Soloha48 [4]

Answer:

hope this helps you

7 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

ASAP PLEAE HELP

Licemer1 [7]

Answer:

21.801

Good luck

5 0

3 years ago