A teacher needs to purchase notebooks and pencils for her students. Notebooks come in packages of 6 and pencils in packages of 10. The table shows the cost of the items. What is the least amount of money the teacher can spend and have the same number of notebooks and pencils.

$31.00

Step-by-step explanation:

Given:

We need to find the LCM of 6 and 10:

Multiples of 6:  6, 12, 18, 24, 30

Multiples of 10: 10, 20, 30

Therefore, the LCM = 30

So the least number of notebooks and pencils the teacher can buy to have the same numbers of each is 30.

⇒ 30 ÷ 6 = 5 packages of notebooks

⇒ 30 ÷ 10 = 3 packages of pencils

5 packages of notebooks = 5 × $5.00 = $25.00

3 packages of pencils = 3 × $2.00 = $6.00

Therefore, the least amount of money the teacher can spend to have the same number of notebooks and pencils (30 each) is:

⇒  $25.00 + $6.00 = $31.00

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5