Ask question

Ask question

All categories

2 years ago

9

Given the following formula,solve for tv=u+at

1 answer:

ad-work [718]2 years ago

3 0

Answer:

t=v-u/a

Step-by-step explanation:

t= v-u/a

we must move u to the other side and divide over a

You might be interested in

What is the vertex of the function?

Romashka [77]

Answer:

(4, -25)

Step-by-step explanation:

One way to answer this is to put the equation into vertex form.

f(x)= x^2 -8x -9 . . . . . given

Add and subtract the square of half the x-coefficient:

f(x) = x^2 -8x +(-8/2)^2 -9 -(-8/2)^2

f(x) = x^2 -8x +16 -25

f(x) = (x -4)^2 -25

Comparing this to the vertex form of a quadratic:

f(x) = a(x -h)^2 +k

we find that (h, k) = (4, -25). This is the vertex.

_____

<em>Alternate solution</em>

The line of symmetry for ...

f(x) = ax^2 +bx +c

is given by x = -b/(2a). For your given quadratic that line is ...

x = -(-8)/(2(1)) = 4

Evaluating f(4) gives the y-coordinate:

f(4) = 4^2 -8·4 -9 = -25

The vertex is (x, y) = (4, -25).

8 0

2 years ago

To create an entry code, you must first choose 2 letters and then, 4 single-digit numbers. How many different entry codes can

Alex_Xolod [135]

Onehundred and forty 2

3 0

2 years ago

Please help, due today

kaheart [24]

Answer:

See below.

Step-by-step explanation:

The domain refers to the x values used in the equation/graph. The line appears to go from -10 to 10 on the x-axis. The dots above -10 and 10 are not shaded in meaning those x values do not count towards the domain meaning we will use parenthesis instead of brackets for interval notation or no equal to signs in inequality notation.

Interval Notation: (-10, 10)

Inequality Notation: -10 < x < 10

4 0

2 years ago

2x+20=-3x-25 help me solve it

zysi [14]

The correct answer is x= -9. Hope this helps!!

8 0

2 years ago

Read 2 more answers

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago