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3 years ago

Solve for the m in the equation

Mathematics

1 answer:

velikii [3]3 years ago

5 0

Answer:

$m = 15$

Step-by-step explanation:

$\frac{m}{9 } + \frac{2}{3} = \frac{7}{3}$

Multiply both sides of the equation by 99, the least common multiple of 9,39,3.

m + 6 =21

Subtract 6 from both sides.

m = 21 - 6

Subtract 6 from 21 to get 15.

m=15

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The height of a ball above the ground as a function of time is given by the function h(t)=−32t2+8t+3 where h is the height of th

gavmur [86]

Answer:

0.125 seconds.

Step-by-step explanation:

We have been given that the height of a ball above the ground as a function of time is given by the function $h(t)=-32t^2+8t+3$ , where h is the height of the ball in feet and t is the time in seconds.

We can see that our given equation is a downward opening parabola as its leading coefficient is negative. The maximum point will be vertex of parabola.

To find the time, when the ball would be at its maximum height, we need to find the x-coordinate of vertex.

Using formula $\frac{-b}{2a}$ , we will find the x-coordinate of vertex of parabola as:

$\frac{-b}{2a}$

$\frac{-8}{2*-32}$

$\frac{-8}{-64}$

$\frac{1}{8}$

$0.125$

Therefore, the ball will be at a maximum height after 0.125 seconds.

6 0

3 years ago

90/9 as a whole number

stiks02 [169]

Answer:

Step-by-step explanation:

5 0

3 years ago

The price of a computer was decreased by 7% to £500. What was the price before the decrease? Give your answer to the nearest pen

aliina [53]

Answer:

537.6

Step-by-step explanation:

0.93*x=500

x=500/0.93

x=537.6

8 0

3 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

3 years ago

What is the value of x ?

zimovet [89]

Each angle of the triangle added together must be equal to 180, so 53+45=98. 180-98=82. So the top angle is 82, the value of x has to be equal to 180-82 which equals 98

8 0

3 years ago