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3 years ago

Why is this equation not valid? Please help

Mathematics

2 answers:

IrinaVladis [17]3 years ago

3 0

Answer:

not valid..

Step-by-step explanation:

wariber [46]3 years ago

3 0

I’m decent at math, but what is this

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HELP ASAP!! HELP FAST!!

erma4kov [3.2K]

The perfect square is answer A

8 0

3 years ago

Convert each measurement 6 quarts = Ounces

nydimaria [60]

Answer:

6 quarts = 192 ounces

Step-by-step explanation:

If one quart is equal to 32 fluid ounces, how much is 6 quarts equal to?

to find out, multiply the number of fluid ounces that one quart has (32) by the number of quarts you want to find in ounces (in this case 6 quarts)

6*32= 192

6 quarts =192 ounces

7 0

4 years ago

PLS HELP WILL GIVE BRAINLY!!

Artist 52 [7]

Answer:

24 units ²

Step-by-step explanation:

In this problem, we are given the circumference of a triangle (after finding the perimeter) and want to find the area of a circle with that circumference. Since the area of a circle is a function based on its radius, we can use the circumference to find the radius to find the area.

First, we can figure out the perimeter of the triangle, which is equal to the sum of its sides. The perimeter is 6+4+7.21 = 17.21 units.

Next, the circumference of a circle is equal to π * diameter = π * 2 * radius. Using 3.14 for π and r for radius, we get

3.14 * 2 * r = 17.21

6.28 * r = 17.21

divide both sides by 6.28 to isolate r

r ≈ 2.74

Furthermore, to find the area from the radius, we can use

area = πr². Plugging 2.74 in for r, we get

2.74² * 3.14 = area

≈23.6, rounding up to 24 units ²

3 0

3 years ago

Koshanah rode her bike for 35 minutes on monday, wendsday, and saterday and 55 minutes each on tusday and thersday . Write an ex

dem82 [27]

Answer:

215 minutes

Step-by-step explanation:

Let

Number of days she rode for 35 minutes = x

Number of days she rode for 55 minutes = y

Total time spent riding her bike = 35x + 55y

x = Monday + Wednesday + Saturday

x = 3 days

y = Tuesday + Thursday

= 2 days

Substitute into the equation

Total time spent riding her bike = 35x + 55y

= 35(3) + 55(2)

= 105 + 110

= 215 minutes

8 0

3 years ago

Consider the two functions:

koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is $x=\frac{40}{4+5a}$

b) The value of the functions at the point where they intersect is $\frac{10 (28 + 15 a)}{4 + 5 a}$

c) The partial derivative of f with respect to $x$ is $\frac{\partial f}{\partial x} = -5a$ and the partial derivative of f with respect to $a$ is $\frac{\partial f}{\partial x} = -5x$

d) The value of $\frac{\partial f}{\partial x}(3,2) = -10$ and $\frac{\partial f}{\partial a}(3,2) = -15$

e) $\upsilon_1=-\frac{3}{4} = -0.75$ and $\upsilon_2=-\frac{3}{4} = -0.75$

f) equation $\upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14}$ and $\upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}$

Step-by-step explanation:

a) In order to find the $x$ we just need to equal the equations and solve for $x$ :

$f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}$

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of $a$ ) must be the same.

$f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}$

and for $g(x)$ :

$g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}$

c) $\frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a$

$\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x$

d) Then evaluating:

$\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10$

$\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15$

e) Substituting the corresponding values:

$\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40} = -\frac{3}{4} = -0.75$

$\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40} = -\frac{3}{4} = -0.75$

f) Writing the equations:

$\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }$

$\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }$

8 0

4 years ago