All categories

3 years ago

BRAINLIEST

Mathematics

1 answer:

mart [117]3 years ago

7 0

Answer:

pretty sure it's b

Step-by-step explanation:

filler filler filler

You might be interested in

The store clerk has 21 mugs to display on shelves. She wants to display a equal number of mugs on each shelf with no mugs left o

elena55 [62]

I believe it’s C, 21 divided by 7. This one doesn’t leave any extra mugs out cause 21 divided by 7 is 3, leaving 7 mugs per 3 shelves and having none left.

8 0

3 years ago

How many 2/3 cup servings are in 1/6 of a cup of popcorn? (in simplest fraction form) A) 1/2 B) 1/4 C) 2/3 D) 3/8

rusak2 [61]

Answer:

1/4 thus B: is your answer

Step-by-step explanation:

Simplify the following:

(1/6)/(2/3)

Multiply the numerator by the reciprocal of the denominator, (1/6)/(2/3) = 1/6×3/2:

3/(6×2)

3/6 = 3/(3×2) = 1/2:

1/(2×2)

2×2 = 4:

Answer: 1/4

7 0

4 years ago

What is<br> 3x-y=9<br> 2x-y=7

svet-max [94.6K]

Answer:

x=2 and y= -3

Step-by-step explanation:

This is a simultaneous equation. To solve this type of equation, there are three methods; substitution, Elimination and Graphical.

But here, we would be using the substitution method.

3x-y=9 Equation 1

2x-y=7. Equation 2

Getting y from equation 2, we have

-y= 7-2x

Multiply both sides by -

y= 2x-7 Equation 3

Substituting y for 2x-7 in equation 1, we have

3x- (2x-7)=9

3x-2x+7=9

x+7=9

x=9-7

x=2

Substituting x as 2 in equation 3

y=2x-7

y= 2(2)-7

y= 4-7

y= -3

5 0

3 years ago

Read 2 more answers

Find all pairs of real numbers $(x,y)$ such that $x + y = 6$ and $x^2 + y^2 = 28$. if you find more than one pair, then list you

vivado [14]

1. $x^2+y^2=28$ and x+y=6

2. x^2+y^2 and x+y remind us of the formula: $(x+y)^{2}= x^2+y^2+2xy$

substituting we get:

$6^{2}= 28+2xy$

2xy=36-28
xy=8/2=4

3 Let's solve the system of equations i)x+y=6 and ii)xy=4

substitute y=4/x in (i):

$x+ \frac{4}{x} =6$

multiply all sides by x:

$x^{2} +4=6x$

$x^{2} -6x+4=0$

complete the square:

$x^{2} -6x+4=x^{2} -2*3x+9-9+4= (x-3)^{2}-5=0$

$x-3=\sqrt{5}$ or $x-3=-\sqrt{5}$

$x=3+\sqrt{5}$ or $x=3-\sqrt{5}$

so from x+y=6, y=6-x

case 1, $x=3+\sqrt{5}$ gives $y=3-\sqrt{5}$

case 2, $x=3-\sqrt{5}$ gives $y=3+\sqrt{5}$

Answer:

$(3-\sqrt{5}, 3+\sqrt{5})$ , $(3+\sqrt{5}, 3-\sqrt{5})$

3 0

3 years ago

Teach me how to ''write each rational number in a form a/b where a and b are integers

Vesna [10]

Write the number over its next next factor of 10. Then multiply by 10s to get rid of the decimal (if there is). Then just simplify.

8 0

3 years ago