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2 years ago

Hello, pls help on these exercises

Mathematics

1 answer:

kvv77 [185]2 years ago

5 0

Answer:

c.p= R.s 75 ,loss=R.s 9 solve it

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Which graph best represents the function f(x)=2x ?

yan [13]

Answer:

Step-by-step explanation:

The best - well, only way- is to check a few points.

Namely -1 (unless your eyesight is really poor!), 0, 1, 2, and 3.

$f(x) = 2^x\\f(-1) = 2^-^1 = \frac12\\f(0) =2^0 =1\\f(1)= 2^1=2\\f(2)=2^2=4\\f(3)=2^3= 8$

Now you can mark all these points in each graph (well, you could if they were on paper and not on a screen) and see which one of the lines passes through all of them. Spoiler alert, it's the B graph.

A represents $2(2^x) = 2^{x+1}$ , B is the one you want, C is $\frac{2^x}2 = 2^{x-1}$ and D looks like $4^x = 2^2^x$

3 0

2 years ago

4/1=12/3,so 12 cups = quarts

REY [17]

Yeah I think so I just took a wild guess and looked at my measurement sheet

5 0

3 years ago

Read 2 more answers

Can you help me find the area of this triangle??

I am Lyosha [343]

A = (1/2)*b*h = (1/2)*8*10 = 40 . . . . units^2

7 0

3 years ago

YU HAVE 1 CASE OF SOAP BARS AND THERE ARE 150 BARS IN A CASE..YU USE 300 BARS OF SOAP PER DAY. HOW MANY CASES DO YOU NEED TO HAV

Fittoniya [83]

If you need 300 bars per day, you need 2 cases per day (2*150 = 300).

Therefore you need 2*7 = 14 cases per week

You can subtract the case you already have leaving you with 13 cases needed

3 0

3 years ago

From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

5 0

3 years ago

Read 2 more answers