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3 years ago

Pls help! will give brainlist!

Mathematics

1 answer:

oksano4ka [1.4K]3 years ago

7 0

Answer: B. 1/6

Step-by-step explanation:

1/2 + 1/3 = 2/12 = 1/6

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8% of what number is 2?

azamat

Answer:

The number is 25

Step-by-step explanation:

Is means equals and of means multiply

8% *N = 2

Change to decimal form

.08 N =2

Divide by .08

.08N/.08 = 2/.08

N =25

5 0

3 years ago

Read 2 more answers

ANSWER BOTH!!!!

laila [671]

Answer:

A. 35 million

B. $1519 increase

$7719 year end

Step-by-step explanation:

6 0

3 years ago

URGENT!!!!! I really suck at these help please I beg of you

taurus [48]

Answer:

Step-by-step explanation:

1) ∠WVZ and ∠VWX are right angles & VX = = WZ Given

2) WV = VW Reflexive property

3) ΔWVZ ≅ ΔVWX HL

4) ∠Z = ∠X CPCT

CPCT --> Corresponding parts of congruent triangle.

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3 years ago

Four elements A,B,C, & D required to produce a drug for ebola virus treatment are to be mixed in a ratio 1:2:4:5 if this inf

adoni [48]

A: 1
B: 2
C: 4
D: 5
Total = 1 + 2 + 4 + 5 = 12

C:total = 4/12 = 1/3

A full circle has 360 deg.

1/3 * 360 deg = 120 deg.

8 0

3 years ago

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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

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

7 0

3 years ago