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nikitadnepr [17]

3 years ago

12

Which of the options below is equal to 5,000? Select ALL that apply.

1 answer:

balu736 [363]3 years ago

7 0

OPTION A ,C AND E IS CORRECT

1/10 x 50,000 = 5,000

100 x 50 =5,000

10 x 500 = 5,000

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❗️❗️❗️Find the length of side x in simplest radical form with a rational denominator.❗️❗️❗️

tiny-mole [99]

Answer:

$x=\frac{\sqrt{14} }{2}$

Step-by-step explanation:

Notice that you are given an isosceles right-angle triangle to solve, since each of its two acute angles measures $45^o$ . Then such means that the sides opposite to these acute angles (the so called "legs" of this right angle triangle) must also be of the same length (x).

We can then use the Pythagorean theorem that relates the square of the hypotenuse to the addition of the squares of the triangles legs:

$(\sqrt{7})^2=x^2+x^2\\7=2\,x^2\\x^2=\frac{7}{2} \\x=+/-\sqrt{\frac{7}{2}} \\x=+/-\frac{\sqrt{14} }{2}$

We use just the positive root, since we are looking for an actual length. then, the requested side is:

$x=\frac{\sqrt{14} }{2}$

6 0

3 years ago

What is the smallest number with the factors 1,2,3,4, and 5? What is the special name for it?

klasskru [66]

To find LCM
factor and eliminate ones already represented
1=1
2=2
3=3
4=2*2
5=5

1*2*2*3*5=60

7 0

3 years ago

2(3x+1)=11 help me plz

padilas [110]

Answer:

x=18

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The cost in dollars (in dollars) of making x flip flops is represented by 40+4x. The cost (in dollars) of making x sandals is re

Anastasy [175]

Answer:

a. C = 12x + 110; b. $290

Step-by-step explanation:

a. The polynomial

Let L = cost of flip-flops

and S = cost of sandals

Then

C = L + S

= 40 + 4x + 70 + 8x

= 12x + 110

b. The cost

C = 12 × 15 + 110

= 180 + 110

= $290

The cost of making 15 flip-flops and 15 sandals is $290.

6 0

3 years ago

Lim x-1 x2 - 1/ sin(x-2)

balu736 [363]

Answer:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0$

Explanation:

Assuming the correct expression is to find the following limit:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}$

Evaluate the numerator:

$\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}$

Evaluate the denominator:

Since $\lim_{x \to1}sin(x-2)\neq 0$

$\frac{0}{\lim_{x \to1}sin(x-2)}=0$

4 0

3 years ago