Please help its iready diagnostic

Ne4ueva [31]

First answer. (: driving is difficult

4 0

4 years ago

if the first 15 sandwiches sold at a restaurant one day, 9 were club sandwiches. at the same rate, how many sandwiches would hav

MaRussiya [10]

Answer:

54 I think

Step-by-step explanation:

I just did 15 divided by 9 and got 5/3 and then did 90 divided by 5/3 and got 54

6 0

3 years ago

Todea what did one pencil say to another mathematics algebra order of operations answer

Serjik [45]

Part 1

$6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)} \\ \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\ \\ =3$

Part 2

$7-8\div4(3^2-1)=7-8\div4(9-1) \\ \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\ \\ =6 \frac{1}{2} = \frac{13}{2}$

Part 3

$2-2(2-5)^2+3^2=2-2(-3)^2+9 \\ \\ =2-2(9)+9=2-18+9=-7$

Part 4

$(5^2-9\cdot3)^2-11=(25-27)^2-11 \\ \\ =(-2)^2-11=4-11=-7$

Part 5

$[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\ \\ =(1-5)\cdot4=-4\cdot4=-16$

Part 6

$12-11\cdot2+16\div8=12-22+2 \\ \\ =-8$

Part 7

$-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\ \\ =-2-(-1)=-2+1=-1$

Part 8

$8+2\cdot3-14\div7=8+6-2 \\ \\ =12$

Part 9

$(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\ \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\ \\ =36+4=40$

Part 10

$7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\ \\ =7+(-7)\cdot5=7+(-35)=7-35 \\ \\ =-28$

Part 11

$3-2[2^3+(-1)^3]=3-2[8+(-1)] \\ \\ =3-2(8-1)=3-2(7)=3-14 \\ \\ -11$

Part 12

$18+6\div3(2^2+5)=18+6\div3(4+5) \\ \\ =18+6\div3(9)=18+6\div27=18+ \frac{2}{9} \\ \\ =18 \frac{2}{9}$

8 0

4 years ago

Find the error in the following sample of work, if one exists.

ch4aika [34]

Answer:

D. There is no mistake.

Step-by-step explanation:

The following lines show the process of factorization by using common factor.

Line 1:

In line 1, the equation is given and is completely fine.

$y^{2}+8xy+2y+16x=0$

The only thing missing was equate to zero, but the options below talk about correct factors only, therefore this can't be considered as a mistake and can be ignored completely.

Line 2:

In line 2, the terms are grouped, from which we can factor out common terms.

$(y^{2}+8xy)+(2y+16x)=0$

This is also fine.

Line 3:

In line 3, the common term y is taken out from group 1 and 2 from other group.

$y(y+8x)+2(y+8x)=0$

which is exactly what is given in line 3.

Line 4:

In line 4 the common factors can be seen and easily split into 2 factors.

$(y+2)(y+8x)=0$

which is exactly what is given in line 4.

Options:

A. The grouping is correct in line 2. So this option is does not hold.

B. Common factor was factored correctly from group 1. So this option does not hold.

C. Common factor was factored correctly from group 2. So this option does not hold.

D. There is no mistake. This is correct. Thus we choose this option as correct answer.

3 0

3 years ago

Record your answers as asum of unitfractions.

Katyanochek1 [597]

1 2/3 = 5/3
12/10 = 6/3

3 0

3 years ago