Ask question

Ask question

All categories

3 years ago

11

Paul's dad made a turkey pot pie for dinner on Wednesday. The family ate4/8 of the pie. On Thursday after school, Paul ate 2/16

of the pie for a snack. What fraction of the pie remained.

2 answers:

valentinak56 [21]3 years ago

8 0

3/8

You need to simplify 2/16

What is a common factor for 2 and 16?

2!

2 divided by 2 is 1
16 divided by 2 is 8

1/8

4/8 - 1/8 = 3/8

3/8

Naya [18.7K]3 years ago

5 0

3/8 please have brainelist, really need

You might be interested in

Which of the following is the equation of a line perpendicular to the line y = -10x + 1, passing through the point (5,7)?

Viktor [21]

Line 1;
y=-10x+1
Gradient m= -10
Gradient of line 2= 1/10
$\frac{y-5}{x-7} = \frac{1}{10}$
y-5= $\frac{1}{10} {x-7}$
y-5= $\frac{1}{10} x- \frac{7}{10}$
y= $\frac{1}{10} x- \frac{43}{10}$

6 0

3 years ago

Read 2 more answers

Use the formula for the difference of cubes to show that 125 - 27 = 98

IrinaVladis [17]

The formula is

a^3 - b^2 = (a - b)(a^2 + ab + b^2) so plugging in the given values:-

a = cube root of 125 = 5 and b = cube root of 27 = 3 , we get:-

125 - 27 = (5 - 3)(5^2 + 5*3 + 3^2)

= 2 * 49

= 98 answer

3 0

3 years ago

Which of the following functions best describes this graph?

jok3333 [9.3K]

<h3>Answer: Choice C</h3>

Explanation:

The x intercepts or roots are x = 3 and x = 5, which lead to the factors x-3 and x-5 respectively.

Multiplying out those factors gets us this:

(x-3)(x-5)

x(x-5)-3(x-5)

x^2-5x-3x+15

x^2-8x+15

8 0

2 years ago

Find the distance between the points (3,‐4) and (3,‐2

balu736 [363]

2 if we’re talking coordinates

8 0

2 years ago

See question in attached photo.

statuscvo [17]

9514 1404 393

Answer:

(a, b) = (-2, -1)

Step-by-step explanation:

The transpose of the given matrix is ...

$A^T=\left[\begin{array}{ccc}1&2&a\\2&1&2\\2&-2&b\end{array}\right]$

Then the [3,1] term of the product is ...

$(A\cdot A^T)_{31}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}1&2&2\end{array}\right]=a+2b+4$

and the [3,2] term is ...

$(A\cdot A^T)_{32}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}2&1&-2\end{array}\right]=2a-2b+2$

Both of these terms in the product matrix are 0. We can solve the system of equations by adding these two terms:

(a +2b +4) +(2a -2b +2) = (0) +(0)

3a +6 = 0

a = -2

Substituting for 'a' in term [3,1] gives ...

-2 +2b +4 = 0

b = -1

The ordered pair (a, b) is (-2, -1).

8 0

2 years ago