Which area on the map benefited MOST from the space program? A) 1 B) 2 C) 3 D) 4 PLSS I NEED A 100!!!!!!! HURRY :[

Mathematics

2 answers:

ozzi3 years ago

5 0

Answer:

wheres the pic

Step-by-step explanation:

romanna [79]3 years ago

3 0

Need more info to confirm answer

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Is the inequality sometimes,always, or never true ?

Paha777 [63]

A. 3(x + 3) > -3(2 + x) SOMETIMES TRUE

3(x + 3) > -3(2 + x) |use distributive property a(b + c) = ab + ac

3x + 9 > -6 - 3x |add 3x to both sides

6x + 9 > -6 |subtract 9 from both sides

6x > -15 |divide both sides by 6

x > -15/6

x > -2.5

B. 9 - x - 5 < - x + 4 NEVER TRUE

9 - x - 5 < - x + 4

4 - x < - x + 4 |add x to both sides

4 < 4 FALSE

3 0

3 years ago

The ratio of adults to children is 5 to 3. If there is 120 people. How many are children?

Lady_Fox [76]

Answer is 72 c.

what you do first is 120/5

= 24

24* num. of children 3 = 72 thuse its the answer

4 0

4 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$