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3 years ago

!!!!!!Pls Help!!!!!!!

Mathematics

2 answers:

seropon [69]3 years ago

6 0

Answer:graph c

Step-by-step explanation:

GuDViN [60]3 years ago

3 0

Answer:

graph c

Step-by-step explanation:

12-5=7 cos she spent $5 of her $12 onda fruit

7/2.5=2.8 therefore itz graph C,, A only goes to 2 and B goes up to 3 which she cannot afford and D is more than she even has so i mean

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What point lies on the line with point slope equation y-3=4(x+7)?

Sidana [21]

Answer:

(-7, 3)

General Formulas and Concepts:

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate
y₁ - y coordinate
m - slope

Step-by-step explanation:

Step 1: Define

Identify

y - 3 = 4(x + 7)

↓ Compare to Point-Slope Form

Point (-7, 3)

Slope m = 4

8 0

3 years ago

How do I solve this using cosine or sine ratio?

skad [1K]

You would use the simple format SOH-CAH-TOA
SOH: sine = opposite/hypotenuse
CAH: cosine = adjacent/hypotenuse
TOA: tangent = opposite/adjacent
----
So for this problem to find c you would use this equation:
sin42= c/7
7sin42=c
c=4.68

Then for d you would use this equation:
tan48=d/4.68
4.68tan48=d
=5.197

Hope this helps :)

6 0

4 years ago

zheka24 [161]

B infinite solutions,
Because both left hand and right hand sides are equal

4 0

4 years ago

Consider the quadratic expression t^2+4t-77. Factor using the box method model.

AnnZ [28]

Answer:

(t−7)(t+11)

Step-by-step explanation:

Let's factor t2+4t−77

t2+4t−77

The middle number is 4 and the last number is -77.

Factoring means we want something like

(t+_)(t+_)

Which numbers go in the blanks?

We need two numbers that...

Add together to get 4

Multiply together to get -77

Can you think of the two numbers?

Try -7 and 11:

-7+11 = 4

-7*11 = -77

Fill in the blanks in

(t+_)(t+_)

with -7 and 11 to get...

(t-7)(t+11)

4 0

3 years ago

A 50-gal tank initially contains 10 gal of fresh water. At t = 0, a brine solution

scZoUnD [109]

$\huge \mathbb{SOLUTION:}$

$\begin{array}{l} \textsf{Let }A(t)\textsf{ be the function which gives the amount} \\ \textsf{of the salt dissolved in the liquid in the tank at} \\ \textsf{any time }t. \textsf{ We want to develop a differential} \\ \textsf{equation that, when solved, will give us an} \\ \textsf{expression for }A(t). \\ \\ \textsf{The basic principle determining the differential} \\ \textsf{equation is} \\ \\ \end{array}$

$\boxed{ \footnotesize \begin{array}{l} \qquad\quad \quad\Large{\dfrac{dA}{dt} = R_{in} - R_{out}} \\ \\ \textsf{where:} \\ \\ \begin{aligned} \bullet\: R_{in} &= \textsf{rate of the salt entering} \\ &= \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt inflow}\end{array}}\right) \times \small(\textsf{Input of brine}) \\ \\ \bullet\: R_{out} &= \textsf{rate of the salt leaving} \\ &= \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt outflow}\end{array}}\right) \times \small(\textsf{Output of brine}) \end{aligned} \end{array}} \\ \\$

$\begin{array}{l} \textsf{On the problem, the amount of salt in the tank,} \\ A(t), \textsf{changes overtime is given by the differential} \\ \textsf{equation} \\ \\ \footnotesize A'(t) = \left(\dfrac{4\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{1\ \textsf{lb}}{1\ \textsf{gal}}\right) - \left(\dfrac{2\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{A(t)\ \textsf{lb}}{10 + (4 - 2)t\ \textsf{gal}}\right) \\ \\ \textsf{There's no salt in the tank (fresh water) at the} \\ \textsf{start, so }A(0) = 0. \textsf{ The amount of solution in the} \\ \textsf{tank is given by }10 + (4 -2)t, \textsf{so the tank will} \\ \textsf{overflow once this expression is equal to the total} \\ \textsf{volume or capacity of the tank.} \\ \\ 10 + (4 - 2)t = 50 \\ \\ \textsf{Solving for }t,\textsf{ we get} \\ \\ \implies \boxed{t = 20\textsf{ mins}} \\ \\ A'(t) = 4 - \dfrac{2A(t)}{10 + 2t} \\ \\ A'(t) = 4 - \dfrac{1}{5 + t} A(t) \\ \\ A'(t) + \dfrac{1}{5 + t} A(t) = 4 \\ \\ \textsf{This is a linear ODE with integrating factor} \\ \mu (t) = e^{\int \frac{1}{5 + t}\ dt} = e^{\ln |5 + t|} = 5 + t \\ \\ \textsf{Multiplying this to the ODE, we get} \\ \\ (5 + t)A'(t) + A(t) = 4(5 + t) \\ \\ [(5 + t)A(t)]' = 20 + 4t \\ \\ (5 + t)A(t) = 20t + 2t^2 + C \\ \\ \textsf{Since }A(0) = 0, \textsf{ we get } C = 0. \\ \\ A(t) = \dfrac{2t^2 + 20t}{t + 5} \\ \\ A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{So the function that gives the amount of salt at} \\ \textsf{any given time }t,\textsf{ is given by} \\ \\ \implies A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{The amount of salt in the tank at the moment} \\ \textsf{of overflow or at }t = 20\textsf{ mins is equal to} \\ \\ A(20) = 2(20) + 10 - \dfrac{50}{20 + 5} \\ \\ \implies \boxed{A = 48\ \textsf{gallons}} \end{array}$

$\Large \mathbb{ANSWER:}$

$\qquad\red{\boxed{\begin{array}{l} \textsf{a. }20\textsf{ mins} \\ \\ \textsf{b. }48\textsf{ gallons}\end{array}}}$

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#5-MinutesAnswer

5 0

3 years ago