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3 years ago

What’s the equation for the table attached

Mathematics

1 answer:

sashaice [31]3 years ago

5 0

Answer:

y = - $\frac{1}{2}x$ + 8

Step-by-step explanation:

So the y-intercept is already given; (0, 8)

So lets calculate the slope; $\frac{y_{2} -y_{1} }{x_{2}-x_{1}}$

$\frac{7-8}{2-0} =-\frac{1}{2}$

The slope is - $\frac{1}{2}$

Let's put it in slope-intercept form;

y = - $\frac{1}{2}x$ + 8

You might be interested in

Express this to single logarithm

zzz [600]

Answer: $\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)$

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

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Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
B*log(A) = log(A^B)

From there, we can then say the following:

$\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}$

8 0

2 years ago

Katie is building a rectangular wall. The wall will be 332 bricks wide and 39 bricks tall. She has 15,000 bricks. Does she have

agasfer [191]

Katie has enough bricks because it only takes 12,948 bricks to build the wall. To get 12,948, you will multiply 332 by 39. So basically you are multiplying the width by how tall it is.

4 0

3 years ago

Read 2 more answers

You and your friend each collect rocks and fossils. Your friend collects three times as many rocks and half as many fossils as y

pochemuha

Answer:

I collect 4 rocks and 16 fossils. While my friend collects 12 rocks and 8 fossils.

Step-by-step explanation:

The number of total objects collected by me = 20

The number of objects collected by my friend = 20

Let us assume:

The number of rocks collected by me = m

The number of fossils collected by me = 2 k

⇒ m + 2 k = 20 .... (1)

So, according to the question:

The number of rocks collected by my friend = 3 x (rocks collected by me)

= 3 x (m) = 3 m

The number of fossils collected by my friend = 1/2 x (fossils collected by me) = 1/2 x (2k ) = k

⇒ 3 m + k = 20 .... (2)

Solving (1)and (2) for the value of m and k, we get:

m + 2 k = 20 ⇒ m = 20 - 2 k ... Put this in (2)

3 m + k = 20 ⇒ 3 (20 - 2 k) + k = 20

⇒ 60 - 6 k + k = 20

or, - 5k = -40

or, k = 40/5 = 8

⇒ k= 8

m + 2(8) = 20

or, m = 4

Hence, the number of rocks collected by me = m = 4

The number of fossils collected by me = 2 k = 2 (4) = 16

Hence, the number of rocks collected by my friend = 3 m = 3 x 4 = 12

The number of fossils collected by my friend = k = 8

Hence, I collect 4 rocks and 16 fossils. While my friend collects 12 rocks and 8 fossils.

4 0

3 years ago

Part F<br> Describe the shape of the graph.

Leokris [45]

This graph is growing exponentially, 0.5 more each time at every point

5 0

2 years ago

(1 point) The parabola given by the equation y=−x2+10x−23 has its vertex at (h,k) for:

Sunny_sXe [5.5K]

the vertex is (5, 2)

5 0

3 years ago