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3 years ago

HELPPPPPP PLEASEEEEE ILL GIVE BRAINLIEST AND POINTS

2 answers:

6 0

1.) The ABC strategy is credited for bringing the HIV/AIDS epidemic under control in Uganda. By promoting abstinence, being faithful, and condom use, safe(r) behaviours have been identified that are applicable to people in different circumstances.

2.) Because the people for whom the propaganda is done are mostly children above 10, who tend to start having sex at an early age.

3.) A reexamination of HIV seroconversion studies suggests that condoms are 90 to 95% effective when used consistently, i.e. consistent condom users are 10 to 20 times less likely to become infected when exposed to the virus than are inconsistent or non-users.

4.) illustrate the photo

Yanka [14]3 years ago

3 0

Answer:

They called the campaign "ABC"--Abstain, Be faithful, or, as a last resort, use a Condom. Within a few years, Uganda had developed what researchers call a "social vaccine" against HIV: cultural norms about sexual responsibility, preached in clinics and public schools, as well as churches and mosques.

Step-by-step explanation:

The ABC strategy is credited for bringing the HIV/AIDS epidemic under control in Uganda. By promoting abstinence, being faithful, and condom use, safe(r) behaviors have been identified that are applicable to people in different circumstances.

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Can someone help me please

vodomira [7]

Answer:

DNE

Step-by-step explanation:

The left limit is +1; the right limit is -1. These are different, so the limit does not exist.

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3 years ago

PLEASE HELP ASAP -- WITH THE WORKK<br><br>50 points

MAXImum [283]

Part-A:-

$\\ \sf\longmapsto x^2+3x-70(4x-12)$

$\\ \sf\longmapsto 4x^2+12x-280-12x^2+36x+840$

$\\ \sf\longmapsto -8x^2+48x+560$

Now

$\\ \sf\longmapsto (f.g)(3)$

$\\ \sf\longmapsto -8(3)^2+48(3)+560$

$\\ \sf\longmapsto -72+144+560$

$\\ \sf\longmapsto 72+560$

$\\ \sf\longmapsto 632$

Part-B:-

$\\ \sf\longmapsto f(3)$

$\\ \sf\longmapsto (3)^2+3(3)-70$

$\\ \sf\longmapsto 9+9-70$

$\\ \sf\longmapsto 18-70$

$\\ \sf\longmapsto -52$

Now

$\\ \sf\longmapsto g(f(3))$

$\\ \sf\longmapsto g(-52)$

$\\ \sf\longmapsto 4(-52-3)$

$\\ \sf\longmapsto 4(-55)$

$\\ \sf\longmapsto -220$

Part:-C

$\\ \sf\longmapsto g(-2)$

$\\ \sf\longmapsto 4(-2-3)$

$\\ \sf\longmapsto 4(-5)$

$\\ \sf\longmapsto -20$

Now

$\\ \sf\longmapsto f(g(-2))$

$\\ \sf\longmapsto f(-20)$

$\\ \sf\longmapsto (-20)^2+3(-20)-70$

$\\ \sf\longmapsto 400-60-70$

$\\ \sf\longmapsto 400-130$

$\\ \sf\longmapsto 270$

Part-D:-

g×f=f×g

$\\ \sf\longmapsto (g.f)(-2)$

$\\ \sf\longmapsto -8(-2)^2+48(-2)+560$

$\\ \sf\longmapsto 32-96+560$

$\\ \sf\longmapsto -64+560$

$\\ \sf\longmapsto 496$

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2 years ago

Solve the equation for y (Y-3)^2=4y-12

Jet001 [13]

<span>(Y-3)^2 = 4y -12 <=> Y^2 - 6Y + 9 = 4y - 12 <=> 4y = Y^2 - 6Y + 21 <=> y = (Y^2 - 6Y + 21)/4. Solving the equation for y means that we have to express y relative to the other quantities in the equation.</span>

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3 years ago

How do you solve the equation y-3=-¼(x+3)

likoan [24]

Because you have two variables and only one equation, you can only solve for approximate answers, and not complete, absolute answers.

<h3><u>You can simplify this equation to x = -4y + 12</u></h3>

y - 3 = -1/4(x + 3)

Distributive property.

y - 3 = -1/4x - 3/4

Multiply all terms by 4.

4y - 12 = -x

Multiply all terms by -1

-4y + 12 = x

8 0

3 years ago

Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”

Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± $\sqrt{1-(y-2)^{2}}$

y = ± $\sqrt{1-x^{2}}+2$

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

(x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± $\sqrt{1-(y-2)^{2}}$

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± $\sqrt{1-x^{2}}$

- Add 2 for both sides

∴ y = ± $\sqrt{1-x^{2}}+2$

∴ y = h(x)

- In the function x = ± $\sqrt{1-(y-2)^{2}}$

∵ $\sqrt{1-(y-2)^{2}}$ ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± $\sqrt{1-x^{2}}+2$

∵ $\sqrt{1-x^{2}}$ ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

8 0

3 years ago