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3 years ago

The length of the two legs of a right triangle are 18 cm and 24 cm what is the length of the hypotenuse of the triangle ?

Mathematics

1 answer:

Setler79 [48]3 years ago

7 0

Answer:

A. 30cm

Step-by-step explanation:

You would need to use the Pythagorean Theorem to find the answer. The theorem is a^2+b^2=c^2.

A and B represent the legs and C represents the hypotenuse.

Now substitute the values: 18^2+24^2=c^2

Calculate the exponents: 324+576=c^2

Now add: 900=c^2

Find the square root of 900: 30

So, the length of the hypotenuse is 30cm.

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Factor the expression by finding the GCF. 16m2 − 12m

ziro4ka [17]

Answer:

4m(4m-3)

Step-by-step explanation:

16m^2 − 12m

16 m^2 = 2*2*2*2*m*m

12m = 2*2*3*m

The greatest factor for 16m^2 and 12m is 2*2*m or 4m

Factor out 4m

16 m^2 = 2*2*2*2*m*m = 4m(2*2m)=4m(4m)

12m = 2*2*3*m = 4m(3)

Factoring out 4m

16m^2 − 12m

4m(4m-3)

3 0

3 years ago

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Math please please please help

kondaur [170]

y=3|3+2|

Hope this helps!

-Payshence

8 0

3 years ago

What is the first quartile of the following data set?<br> 15, 18, 20, 21, 23, 24, 26, 29, 34, 37, 40

nikklg [1K]

Answer:

20.5

Step-by-step explanation:

20 + 21 = 41. Divide 41 by 2 than you get 20.5.

3 0

3 years ago

In problem solve the given differential equation by underdetermined coefficients y''-2y+y=xe^x

Alexus [3.1K]

Answer:

Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$

Step-by-step explanation:

Given Differential Equation,

$y"-2y'+y=xe^x$ ...............(1)

We need to solve the given differential equations using undetermined coefficients.

Let the solution of the given differential equation is made up of two parts. one complimentary solution and one is particular solution.

$\implies\:y(x)=y_c(x)+y_p(x)$

For Complimentary solution,

Auxiliary equation is as follows

m² - 2m + 1 = 0

( m - 1 )² = 0

m = 1 , 1

So,

$y_c(x)=C_1e^x+c_2xe^x$

Now for particular solution,

let $y_p(x)=Ax^3e^x$

$y'=Ax^3e^x+3Ax^2e^x$

$y"=Ax^3e^x+6Ax^2e^x+6Axe^x$

Now putting these values in (1), we get

$Ax^3e^x+6Ae^2e^x+6Axe^x-2(Ax^3e^x+3Ax^2e^x)+Ax^3e^x=xe^x$

$Ax^3e^x+6Ae^2e^x+6Axe^x-2Ax^3e^x-6Ax^2e^x+Ax^3e^x=xe^x$

$6Axe^x=xe^x$

$6A=1$

$A=\frac{1}{6}$

$\implies\:y_p(x)=\frac{x^3e^x}{6}$

Therefore, Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$

8 0

3 years ago

-1≤ 3x-10≤ 2 please help me

kicyunya [14]

Answer:

3 ≤ x ≤ 4

Step-by-step explanation:

Step 1: Add 10 to all parts/sections.

$-1 + 10 \leq 3x - 10 + 10 \leq 2+10$
$9\leq 3x\leq 12$

Step 2: Divide all parts/sections by 3.

$\frac{9}{3} \leq \frac{3x}{3} \leq \frac{12}{3}$
$3 \leq x \leq 4$

Therefore, the answer is 3 ≤ x ≤ 4.

5 0

2 years ago

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