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3 years ago

The difference between three times a number and 15 is equal to 6. Find the number

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

5 0

15+6=21
21÷3=7
answer
3×7-15=6

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Graph from liner standard form 3x+2y=12?

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Answer:

See picture below.

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2 years ago

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Hi can someone please help me with this i will give you five stars!!

tino4ka555 [31]

Step-by-step explanation:

Let's look at what we know. We know that...

P=2000

r=0.04 (Change 4% to a decimal)

t=7 (25 years minus 18 years equals 7)

n=1

Since we are compounding each year, we need to use this equation: $P(1+\frac{r}{n} )^{nt}$

Now just plug the numbers in: (Answer to #1:) $2000(1+\frac{0.04}{1} )^{(1*7)}$

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When we round, this equals to $2,632. (Answer to #2).

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2 years ago

Find the gradient (m) and y-intercept (c) of the equation y =-2/3x+1/3

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he equation of a line is written as y=mx+b where m is the slope and b is the

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2 years ago

A company has 14 employees with a salary of $21,000, 11 employees with a salary of $22,800, 17 employees with a salary of $

kodGreya [7K]

Answer:

27525.4902

Step-by-step explanation:

First need to add the total number of employees:

14+11+17+3+5+1=51

Now multiply the salary for each number of employees then add to have the total:

14x21000+11x22800+17x24800+3x32200+5x38500+1x148300= 1403800

Now to obtain the mean value just divide the total salary by the total number of employees:

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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.

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3 years ago

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