Identify a number that results in a 4-digit-number when 156,350 is subtracted from it

A company makes steel rods shaped like cylinders. Each rod has a radius of 4 centimeters and a height of 20 centimeters. If the

Tamiku [17]

(One steel rod’s volume) = 3.14 x 4^2 x 20

= 1004.8cm^3

(Used steel) / (One steel rod’s volume) =

(Made steel rod’s number)

The answer is 28

5 0

2 years ago

The ampitude/frequency/period of y=1/4cos1/2x is 4 pie

Fantom [35]

Amplitude: 1/4 (= |a|)
Frequency: 1/2 (= |b|)
Period: 4pi (= 2pi/|b|)

7 0

3 years ago

Last year there were 120 students in choir. This year, 30% more students took choir. How many students are taking choir this yea

soldi70 [24.7K]

Answer:

156 students

Step-by-step explanation:

To find how many students are in choir this year, multiply 120 by 1.3

120(1.3)

= 156

So, this year, 156 students are taking choir

7 0

3 years ago

Read 2 more answers

Good Morning Ethansledd17!<br> what is 54873578x958498=?

pickupchik [31]

Answer: 5.259621477e13

Step-by-step explanation: Hoped this helped!!!

5 0

2 years ago

Read 2 more answers

Can someone please help me with my maths question

DIA [1.3K]

Answer:

$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$

$x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

4 0

3 years ago