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3 years ago

10

A block of stone measures 15 cm X 15 cm X 20 cm. What is the total surface area of the stone?(Hint: A block has six sides.)

1 answer:

BARSIC [14]3 years ago

5 0

$Look\ at\ the\ picture.\\\\2\ sides:\ 15cm\times15cm\\\\Area:A_1=2\times(15cm\times15cm)=2\times225cm^2=450cm^2\\\\4\ sides:15cm\times20cm\\\\Area:A_2=4\times(15cm\times20cm)=4\times300cm^2=1200cm^2\\\\Area\ of\ the\ block:\\\\A=A_1+A_2\\\\\boxed{A=450cm^2+1200cm^2=1650cm^2}$

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Which point would be a solution to the system of linear inequalities shown below? y>4 x-5\hspace{50px}y\ge\frac{3}{5} x-1 y&g

Sauron [17]

Answer:

$x < \frac{20}{17}$ and $y \ge \frac{-5}{17}$

Step-by-step explanation:

Given

$y>4 x-5\hspace{50px}y\ge\frac{3}{5} x-1$

Required

Find x and y

In the second equation. Assume that:

$y = \frac{3}{5}x - 1\\$

Substitute $y = \frac{3}{5}x - 1$ in the first equation

$y > 4x - 5$

$\frac{3}{5}x - 1 > 4x - 5$

Collect like terms

$\frac{3}{5}x - 4x > - 5 + 1$

$\frac{3}{5}x - 4x > -4$

Multiply through by 5

$3x - 20x > -20$

$-17x > -20$

Solve for x

$x < \frac{20}{17}$

Substitute this value of x in $y \ge \frac{3}{5}x - 1$

$y \ge \frac{3}{5}*\frac{20}{17} - 1$

$y \ge \frac{3}{1}*\frac{4}{17} - 1$

$y \ge \frac{12}{17} - 1$

$y \ge \frac{12- 17}{17}$

$y \ge \frac{-5}{17}$

3 0

4 years ago

The sum of three consecutive even integers is 36. What is the second integers?

podryga [215]

let the consecutive even integers be 2x , 2x + 2 and 2x + 4

let's solve for values of the integers :

$2x + 2x + 2 + 2x + 4 = 36$

$6x + 6 = 36$

$6x = 36 - 6$

$x = \dfrac{30}{6}$

$x = 5$

therefore, the consecutive even integers are :

$2x = 2 \times 5 = 10$

$2x + 2 = 10 + 2 = 12$

$2x + 4 = 10 + 4 = 14$

I hope it helped...

7 0

3 years ago

How do i solve with two variables

Verdich [7]

50 divided by 13 on both sides
13y/13 =0 50/13 = 3.846153....
change it to y= 3 since 4 is to big.
3•13 = 39
50-39=11
x=11

Final answer
x=11
y=3

6 0

3 years ago

A new bank customer with $4 comma 500 wants to open a money market account. The bank is offering a simple interest rate of 1.5

Jet001 [13]

a) Simple interest earned by customer in 10 years is $ 675

b) Account balance after 10 years is $ 5175

<em><u>Solution:</u></em>

<em><u>The simple interest is given by formula:</u></em>

$Simple\ Interest = \frac{p \times n \times r}{100}$

Where,

"p" is the principal

"r" is the rate of interest

"n" is the number of years

<em><u>a. How much interest will the customer earn in 10 years?</u></em>

From given,

Principal = p = $ 4500

Simple interest rate = r =1.5 %

n = 10

Substituting the values in formula,

$Simple\ Interest = \frac{4500 \times 10 \times 1.5}{100}\\\\Simple\ Interest = 450 \times 1.5\\\\Simple\ Interest = 675$

Thus simple interest earned by customer in 10 years is $ 675

<em><u>b. What will the account balance be after 10 years ?</u></em>

Account balance = principal + simple interest

Account balance = 4500 + 675 = 5175

Thus account balance after 10 years is $ 5175

8 0

3 years ago

NEED HELP !!!!!! READDD THE QUESTION THERE ARE TWO PARTS

mel-nik [20]

We have to find horizontal component of Mayosons velocity.

#a

$\\ \tt\bull\leadsto V_x=vsin\theta$

$\\ \tt\bull\leadsto V_x=2.5sin36$

$\\ \tt\bull\leadsto V_x=2.5(0.58)$

$\\ \tt\bull\leadsto V_x=1.45m/s$

#b

Here 2 vectors A=8 and B=2.5

$\\ \tt\bull\leadsto \overrightarrow{R}$

$\\ \tt\bull\leadsto \sqrt{A^2+B^2+2ABcos\Theta}$

$\\ \tt\bull\leadsto \sqrt{8^2+2.5^2+2(8)(2.5)cos36}$

$\\ \tt\bull\leadsto \sqrt{64+6.25+40(0.8)}$

$\\ \tt\bull\leadsto \sqrt{70.25+32}$

$\\ \tt\bull\leadsto \sqrt{102.25}$

$\\ \tt\bull\leadsto 10.15m/s$

3 0

3 years ago