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3 years ago

Please look for the question in the picture.

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

8 0

Answer:

I beleive the missing value is -3

Step-by-step explanation:

if you plug -3 into the pink box it will give you the result of -5

(anyone is free to correct me)

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9. The perimeter of a rectangle iş 24 meters. The width is 30 meters less than five

serious [3.7K]

Answer:

length of rectangle=7 metre

width of rectangle=5 metre

Step-by-step explanation:

perimeter of rectangle=2(length+width)

given perimeter of rectangle=p=24

let, length of rectangle=l

width of rectangle=b=5l-30

ac/ques, p=2(l+b)

24=2(l+5l-30)

12=6l-30

42=6l

l= $\frac{42}{6}$

l=7 metre

b=35-30=5 metre

5 0

3 years ago

F(x) = \sqrt{x+8} ; g(x) = 8x - 12<br> Find f(g(x)).

mestny [16]

$\mathbf{Task.} ~ f(x)=\sqrt{x+8}; ~ g(x)=8x-12. ~ \mathrm{Find} ~ f(g(x)).$

$f(g(x))=\sqrt{g(x)+8}=\sqrt{8x-12 + 8}=\sqrt{8x-4}=$

$=\sqrt{4(2x-1)}=2\sqrt{2x-1}.$

$Answer \colon f(g(x))=2\sqrt{2x-1}. ~ \blacktriangle$

7 0

3 years ago

At a local grocery, 26-ounces of a sport drink costs $1.30. What is the price per ounce?

anygoal [31]

Your answer would be 0.05 per ounce

5 0

2 years ago

Which of the following inequalities has the graphed situation below ?

alexgriva [62]

Answer:

i need pointsss

Step-by-step explanation:

7 0

3 years ago

An algorithm takes 0.5 seconds to run on an input of size 100. How long will it take to run on an input of size 1000 if the algo

dexar [7]

Answer:

linear: 5s

quadratic: 50s

log-linear: 0.75 s

cubic: 500s

Step-by-step explanation:

Let $t_1,t_2$ be the running time associated with the input of sizes $s_1,s_2$

If the running time is linear

$t_2 = t_1\frac{s_2}{s_1} = 0.5*\frac{1000}{100} = 0.5*10 = 5s$

If the running time is quadratic

$t_2 = t_1\left(\frac{s_2}{s_1}\right)^2 = 0.5*\left(\frac{1000}{100}\right)^2 = 0.5*10^2 = 50s$

If the running time is log-linear

$t_2 = t_1\frac{log(s_2)}{log(s_1)} = 0.5*\frac{log(1000)}{log(100)} = 0.5*1.5 = 0.75s$

If the running time is cubic:

$t_2 = t_1\left(\frac{s_2}{s_1}\right)^3 = 0.5*\left(\frac{1000}{100}\right)^3 = 0.5*10^3 = 500s$

6 0

3 years ago