All categories

3 years ago

Find x,y and 2. (leave answer in simplest radical form) *

Mathematics

1 answer:

True [87]3 years ago

3 0

Answer:

x = 32

y = 16sqrt(5)

z = 8sqrt(5)

Step-by-step explanation:

x/16 = 16/8

x/16 = 2

x = 32

y^2 = 16^2 + 32^2

y^2 = 1280

y = 16sqrt(5)

z^2 = 8^2 + 16^2

z^2 = 320

z = 8sqrt(5)

You might be interested in

6(x-2)=5x+3x+8 solve for x.

Mnenie [13.5K]

Answer:

x = -10

Step-by-step explanation:

6 • (x - 2) - (8x + 8) = 0

Pulling out like terms :

3.1 Pull out like factors :

-2x - 20 = -2 • (x + 10)

Equation at the end of step 3 :

-2 • (x + 10) = 0

Equations which are never true :

4.1 Solve : -2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

4.2 Solve : x+10 = 0

Subtract 10 from both sides of the equation :

x = -10

One solution was found :

x = -10

Processing ends successfully

plz mark me as brainliest :)

4 0

3 years ago

Read 2 more answers

In 2007, there were more than 8.14 million cars for sale. Over the next 3 years, the number of cars decreased by 23%. Write an e

Nina [5.8K]

Answer:

c = 8.14 million×(0.9166)^t

4.83 million

Step-by-step explanation:

Data:

t = y - 2007

c₀ = 8.14 million

c₃ = 23 % less than c₁

Part 1. Calculate c₃

c₃ = c₀(1 - 0.23) = 0.77c₀

Part 2. Calculate r

c₃ = c₀r^t

0.77c₀ = c₀r³

0.77 = r³ Divided each side by c₀

r = 0.9166 Took the cube root of each side

The explicit decay model is c = 8.14 million×(0.9166)^t

Part 3. Prediction

t = 2013 - 2007 = 6

c = c₀r^t = 8.14 million×(0.9166)⁶ = 8.14 million × 0.5929 = 4.83 million

The model predicts that there will be 4.83 million cars for sale in 2013.

7 0

3 years ago

What is the domain of f(x)=5x^2 2x-1

adoni [48]

Domain of 5x^2 + 2x - 1 is all real numbers.

5 0

3 years ago

How to expand the number 2,937,082

denis-greek [22]

2x1,000,000
9x100,000
3x10,000
7x1,000
0x100
8x10
2x1

7 0

3 years ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.

4 0

3 years ago