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valentina_108 [34]

2 years ago

5

At a skills competition, a target is being lifted into the air by a cable at a constant speed. An archer standing on the ground

launches an arrow toward the target. The system of equations below models the height, in feet, of the target and the arrow t seconds after it was fired. Which statement most likely describes the situation modeled by this system? mc016-1.jpg

1 answer:

adell [148]2 years ago

5 0

From the kinematic equations, s(t) = s_in + (v_in)(t) - (1/2)(g)(t^2). This means that there is an initial distance of 4, initial velocity and acceleration (in opposite direction of the motion) f 32. Option 4 shows the correct answer.

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Convert 778.5 million km to scientific notation.

Doss [256]

778.5 Million in Scientific Notation is..

7.785 × $10^{8}$

5 0

2 years ago

the trinomial x2 bx c factors to (x m)(x n). if b is negative and c is positive, what must be true about m and n?

galina1969 [7]

For c to be positive, and for b to be negative, m must be negative and n must be negative.

X^2 - bx + c = (x - m)(x - n).

c is the product of m and n. If both m and n are positive, c would be positive. However b is the sum of m and n, therefore to make b negative, both m and n must be negative to ensure that the product of m and n is positive

7 0

2 years ago

Read 2 more answers

Problem number 26 of the Rhind Papyrus says: Find a quantity such that when it is added to StartFraction 1 Over 4 EndFraction of

STALIN [3.7K]

Answer:

The quantity is 12

Step-by-step explanation:

Here, we want to find the solution to the equation.

Let the quantity we are looking for be x;

According to the question;

The quantity plus a quarter of the quantity = 15

x + 1/4(x) = 15

x + x/4 = 15

Multiply through by 4

4x + (x/4)*4 = 15 * 4

4x + x = 60

5x = 60

x = 60/5

x = 12

5 0

2 years ago

Read 2 more answers

lbvjy [14]

Answer: zgfbnzbzdbzdfbgzbnbgfgfgfbv bzgfbvb

Step-by-step explanation:

8 0

2 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

6 0

2 years ago