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3 years ago

√x+4=√3x A.) x=2 B.) x=-2 C.) x=3 D.) x=-3

Mathematics

2 answers:

Agata [3.3K]3 years ago

6 0

I hope this helps you

choli [55]3 years ago

3 0

The anwser is (A) x=2

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MULTIPLE CHOICE A triangle has an angle that measures 41° and an angle

m_a_m_a [10]

Answer:

117

Step-by-step explanation:

Sum of all the angles of a triangle is 180

Let the third angle be 'x'

41 + 22 + x = 180

63 + x = 180

x = 180 - 63

x = 117

7 0

3 years ago

Read 2 more answers

Help me plz asap i need the first question dont answer till u see file

nika2105 [10]

Okay, so with the link we can tell this:

flour and baking soda are equal ratios
salt is one half the ratio of flour/baking soda

A: correct
flour and baking soda as equal ratios

B: incorrect
salt is at an incorrect ratio to flour

C: correct
salt is at a correct ratio to baking soda

D: incorrect
salt is at an incorrect ratio to baking soda

E: correct
baking soda is at a correct ratio to salt

answers: A, C , E

8 0

3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$